$f: G to mathbb{C}^*$ is a homomorphism. Show that the sum $sum f (g) = 0$ or $n$












5












$begingroup$


Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    What's the order of $i$ then?
    $endgroup$
    – the_fox
    Jan 4 at 3:11






  • 2




    $begingroup$
    Does G have to have finite order?
    $endgroup$
    – Joel Pereira
    Jan 4 at 3:39










  • $begingroup$
    @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    $endgroup$
    – Robert Lewis
    Jan 4 at 3:45










  • $begingroup$
    @the_fox :( back to square one.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    $begingroup$
    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    $endgroup$
    – Joel Pereira
    Jan 4 at 15:33
















5












$begingroup$


Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    What's the order of $i$ then?
    $endgroup$
    – the_fox
    Jan 4 at 3:11






  • 2




    $begingroup$
    Does G have to have finite order?
    $endgroup$
    – Joel Pereira
    Jan 4 at 3:39










  • $begingroup$
    @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    $endgroup$
    – Robert Lewis
    Jan 4 at 3:45










  • $begingroup$
    @the_fox :( back to square one.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    $begingroup$
    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    $endgroup$
    – Joel Pereira
    Jan 4 at 15:33














5












5








5


2



$begingroup$


Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question











$endgroup$




Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism




abstract-algebra group-theory proof-verification finite-groups group-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 11:44







Subhasis Biswas

















asked Jan 4 at 2:51









Subhasis BiswasSubhasis Biswas

512412




512412








  • 6




    $begingroup$
    What's the order of $i$ then?
    $endgroup$
    – the_fox
    Jan 4 at 3:11






  • 2




    $begingroup$
    Does G have to have finite order?
    $endgroup$
    – Joel Pereira
    Jan 4 at 3:39










  • $begingroup$
    @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    $endgroup$
    – Robert Lewis
    Jan 4 at 3:45










  • $begingroup$
    @the_fox :( back to square one.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    $begingroup$
    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    $endgroup$
    – Joel Pereira
    Jan 4 at 15:33














  • 6




    $begingroup$
    What's the order of $i$ then?
    $endgroup$
    – the_fox
    Jan 4 at 3:11






  • 2




    $begingroup$
    Does G have to have finite order?
    $endgroup$
    – Joel Pereira
    Jan 4 at 3:39










  • $begingroup$
    @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    $endgroup$
    – Robert Lewis
    Jan 4 at 3:45










  • $begingroup$
    @the_fox :( back to square one.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    $begingroup$
    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    $endgroup$
    – Joel Pereira
    Jan 4 at 15:33








6




6




$begingroup$
What's the order of $i$ then?
$endgroup$
– the_fox
Jan 4 at 3:11




$begingroup$
What's the order of $i$ then?
$endgroup$
– the_fox
Jan 4 at 3:11




2




2




$begingroup$
Does G have to have finite order?
$endgroup$
– Joel Pereira
Jan 4 at 3:39




$begingroup$
Does G have to have finite order?
$endgroup$
– Joel Pereira
Jan 4 at 3:39












$begingroup$
@JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
$endgroup$
– Robert Lewis
Jan 4 at 3:45




$begingroup$
@JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
$endgroup$
– Robert Lewis
Jan 4 at 3:45












$begingroup$
@the_fox :( back to square one.
$endgroup$
– Subhasis Biswas
Jan 4 at 11:41




$begingroup$
@the_fox :( back to square one.
$endgroup$
– Subhasis Biswas
Jan 4 at 11:41




1




1




$begingroup$
@RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
$endgroup$
– Joel Pereira
Jan 4 at 15:33




$begingroup$
@RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
$endgroup$
– Joel Pereira
Jan 4 at 15:33










2 Answers
2






active

oldest

votes


















6












$begingroup$

It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41



















0












$begingroup$

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061265%2ff-g-to-mathbbc-is-a-homomorphism-show-that-the-sum-sum-f-g-0-or%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41
















6












$begingroup$

It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41














6












6








6





$begingroup$

It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer











$endgroup$



It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 6:13









Shaun

10.1k113685




10.1k113685










answered Jan 4 at 3:58









Robert LewisRobert Lewis

48.6k23167




48.6k23167








  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41














  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41








1




1




$begingroup$
@Shaun: nice edit, thanks!
$endgroup$
– Robert Lewis
Jan 4 at 6:23




$begingroup$
@Shaun: nice edit, thanks!
$endgroup$
– Robert Lewis
Jan 4 at 6:23




1




1




$begingroup$
Amazing. Just amazing.
$endgroup$
– Subhasis Biswas
Jan 4 at 11:40




$begingroup$
Amazing. Just amazing.
$endgroup$
– Subhasis Biswas
Jan 4 at 11:40












$begingroup$
Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
$endgroup$
– Cameron Williams
Jan 4 at 12:03




$begingroup$
Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
$endgroup$
– Cameron Williams
Jan 4 at 12:03












$begingroup$
@SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 0:41




$begingroup$
@SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 0:41











0












$begingroup$

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40
















0












$begingroup$

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40














0












0








0





$begingroup$

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer









$endgroup$



Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 4:16









RileyRiley

1825




1825












  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40


















  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40
















$begingroup$
Can you please verify mine?
$endgroup$
– Subhasis Biswas
Jan 4 at 6:00




$begingroup$
Can you please verify mine?
$endgroup$
– Subhasis Biswas
Jan 4 at 6:00












$begingroup$
the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
$endgroup$
– Riley
Jan 4 at 7:01






$begingroup$
the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
$endgroup$
– Riley
Jan 4 at 7:01














$begingroup$
$z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
$endgroup$
– Subhasis Biswas
Jan 4 at 8:27




$begingroup$
$z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
$endgroup$
– Subhasis Biswas
Jan 4 at 8:27












$begingroup$
I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
$endgroup$
– Riley
Jan 4 at 10:37




$begingroup$
I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
$endgroup$
– Riley
Jan 4 at 10:37












$begingroup$
I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
$endgroup$
– Subhasis Biswas
Jan 4 at 11:40




$begingroup$
I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
$endgroup$
– Subhasis Biswas
Jan 4 at 11:40


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061265%2ff-g-to-mathbbc-is-a-homomorphism-show-that-the-sum-sum-f-g-0-or%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How do I know what Microsoft account the skydrive app is syncing to?

Image
up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
Read more

When does type information flow backwards in C++?

Image
up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
Read more

Grease: Live!

Image
A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
Read more