Abelian group G is of order $240$, $sigma(g)=g^{16}in text{Aut}(G)$, then #{ker $sigma$} =?
$begingroup$
I'm told to determine if this is true:
If $G$ is an abelian group of order $240$. Then the kernel of the map $Gto G$ given by $sigma(g)=g^{16}$ has 16 elements.
Since the group is not determined and there are $5$ abelian groups of order $240$ up to isomorphism. They are $mathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ ,
$mathbb{Z}_4bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_8bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_{16}bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_4bigotimesmathbb{Z}_4bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$.
I'm not sure if I need to discuss each case.
abstract-algebra group-theory
edited Dec 8 '18 at 12:15
Andrews
3901317
asked Dec 8 '18 at 10:58
LOISLOIS
3838
$endgroup$
add a comment |
$begingroup$
I'm told to determine if this is true:
If $G$ is an abelian group of order $240$. Then the kernel of the map $Gto G$ given by $sigma(g)=g^{16}$ has 16 elements.
Since the group is not determined and there are $5$ abelian groups of order $240$ up to isomorphism. They are $mathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ ,
$mathbb{Z}_4bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_8bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_{16}bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_4bigotimesmathbb{Z}_4bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$.
I'm not sure if I need to discuss each case.
abstract-algebra group-theory
edited Dec 8 '18 at 12:15
Andrews
3901317
asked Dec 8 '18 at 10:58
LOISLOIS
3838
$endgroup$
1
$begingroup$
Hint: Start by writing $G$ as a direct sum of a $2$-group and the rest.
$endgroup$
– Michael Burr
Dec 8 '18 at 11:04
add a comment |
$begingroup$
I'm told to determine if this is true:
If $G$ is an abelian group of order $240$. Then the kernel of the map $Gto G$ given by $sigma(g)=g^{16}$ has 16 elements.
Since the group is not determined and there are $5$ abelian groups of order $240$ up to isomorphism. They are $mathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ ,
$mathbb{Z}_4bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_8bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_{16}bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_4bigotimesmathbb{Z}_4bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$.
I'm not sure if I need to discuss each case.
abstract-algebra group-theory
edited Dec 8 '18 at 12:15
Andrews
3901317
asked Dec 8 '18 at 10:58
LOISLOIS
3838
$endgroup$
I'm told to determine if this is true:
If $G$ is an abelian group of order $240$. Then the kernel of the map $Gto G$ given by $sigma(g)=g^{16}$ has 16 elements.
Since the group is not determined and there are $5$ abelian groups of order $240$ up to isomorphism. They are $mathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ ,
$mathbb{Z}_4bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_8bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_{16}bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$,
$mathbb{Z}_4bigotimesmathbb{Z}_4bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$.
I'm not sure if I need to discuss each case.
abstract-algebra group-theory
abstract-algebra group-theory
edited Dec 8 '18 at 12:15
Andrews
3901317
asked Dec 8 '18 at 10:58
LOISLOIS
3838
edited Dec 8 '18 at 12:15
Andrews
3901317
asked Dec 8 '18 at 10:58
LOISLOIS
3838
edited Dec 8 '18 at 12:15
Andrews
3901317
edited Dec 8 '18 at 12:15
Andrews
3901317
edited Dec 8 '18 at 12:15
Andrews
3901317
3901317
asked Dec 8 '18 at 10:58
LOISLOIS
3838
asked Dec 8 '18 at 10:58
LOISLOIS
3838
asked Dec 8 '18 at 10:58
LOISLOIS
3838
3838
1
$begingroup$
Hint: Start by writing $G$ as a direct sum of a $2$-group and the rest.
$endgroup$
– Michael Burr
Dec 8 '18 at 11:04
add a comment |
1
$begingroup$
Hint: Start by writing $G$ as a direct sum of a $2$-group and the rest.
$endgroup$
– Michael Burr
Dec 8 '18 at 11:04
1
1
$begingroup$
Hint: Start by writing $G$ as a direct sum of a $2$-group and the rest.
$endgroup$
– Michael Burr
Dec 8 '18 at 11:04
$begingroup$
Hint: Start by writing $G$ as a direct sum of a $2$-group and the rest.
$endgroup$
– Michael Burr
Dec 8 '18 at 11:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Following your approach, you can combine all cases to write
$$
G=Htimes mathbb{Z}/15,
$$
where $H$ is an abelian group of order $16$. An arbitrary element of $G$ can be written as
$$
g=h+k.
$$
Then, since $G$ is abelian,
$$
sigma(g)=sigma(h)+sigma(k).
$$
Use Lagrange's theorem to study $sigma(h)$ and Fermat's little theorem or Lagrange's theorem to study $sigma(k)$.
answered Dec 8 '18 at 11:10
Michael BurrMichael Burr
26.8k23262
$endgroup$
$begingroup$
$G=Htimes mathbb{Z}_{15}$, for$gin G$ ,$g=hcdot k,sigma(g)=(hk)^{16}=h^{16}k^{16}=sigma(g)cdot sigma(k)$, $h in H$ and $H$ is of order 16 hence $sigma(h)=e$ while $k in mathbb{Z}_{15}$ hence $sigma(k)=k^{16}=k$. Therefore $sigma(g)=sigma(k)=k $. $Ker={g=(h,e_k),hin H, e_k in Z_{15}$ is the identity in$ Z_{15}}$ Hence $|Ker(sigma)|=16$. Right?
$endgroup$
– LOIS
Dec 8 '18 at 13:03
$begingroup$
Yes, that's how to finish it.
$endgroup$
– Michael Burr
Dec 8 '18 at 19:16
add a comment |
Your Answer
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$begingroup$
Following your approach, you can combine all cases to write
$$
G=Htimes mathbb{Z}/15,
$$
where $H$ is an abelian group of order $16$. An arbitrary element of $G$ can be written as
$$
g=h+k.
$$
Then, since $G$ is abelian,
$$
sigma(g)=sigma(h)+sigma(k).
$$
Use Lagrange's theorem to study $sigma(h)$ and Fermat's little theorem or Lagrange's theorem to study $sigma(k)$.
answered Dec 8 '18 at 11:10
Michael BurrMichael Burr
26.8k23262
$endgroup$
$begingroup$
$G=Htimes mathbb{Z}_{15}$, for$gin G$ ,$g=hcdot k,sigma(g)=(hk)^{16}=h^{16}k^{16}=sigma(g)cdot sigma(k)$, $h in H$ and $H$ is of order 16 hence $sigma(h)=e$ while $k in mathbb{Z}_{15}$ hence $sigma(k)=k^{16}=k$. Therefore $sigma(g)=sigma(k)=k $. $Ker={g=(h,e_k),hin H, e_k in Z_{15}$ is the identity in$ Z_{15}}$ Hence $|Ker(sigma)|=16$. Right?
$endgroup$
– LOIS
Dec 8 '18 at 13:03
$begingroup$
Yes, that's how to finish it.
$endgroup$
– Michael Burr
Dec 8 '18 at 19:16
add a comment |
$begingroup$
Following your approach, you can combine all cases to write
$$
G=Htimes mathbb{Z}/15,
$$
where $H$ is an abelian group of order $16$. An arbitrary element of $G$ can be written as
$$
g=h+k.
$$
Then, since $G$ is abelian,
$$
sigma(g)=sigma(h)+sigma(k).
$$
Use Lagrange's theorem to study $sigma(h)$ and Fermat's little theorem or Lagrange's theorem to study $sigma(k)$.
answered Dec 8 '18 at 11:10
Michael BurrMichael Burr
26.8k23262
$endgroup$
$begingroup$
$G=Htimes mathbb{Z}_{15}$, for$gin G$ ,$g=hcdot k,sigma(g)=(hk)^{16}=h^{16}k^{16}=sigma(g)cdot sigma(k)$, $h in H$ and $H$ is of order 16 hence $sigma(h)=e$ while $k in mathbb{Z}_{15}$ hence $sigma(k)=k^{16}=k$. Therefore $sigma(g)=sigma(k)=k $. $Ker={g=(h,e_k),hin H, e_k in Z_{15}$ is the identity in$ Z_{15}}$ Hence $|Ker(sigma)|=16$. Right?
$endgroup$
– LOIS
Dec 8 '18 at 13:03
$begingroup$
Yes, that's how to finish it.
$endgroup$
– Michael Burr
Dec 8 '18 at 19:16
add a comment |
$begingroup$
Following your approach, you can combine all cases to write
$$
G=Htimes mathbb{Z}/15,
$$
where $H$ is an abelian group of order $16$. An arbitrary element of $G$ can be written as
$$
g=h+k.
$$
Then, since $G$ is abelian,
$$
sigma(g)=sigma(h)+sigma(k).
$$
Use Lagrange's theorem to study $sigma(h)$ and Fermat's little theorem or Lagrange's theorem to study $sigma(k)$.
answered Dec 8 '18 at 11:10
Michael BurrMichael Burr
26.8k23262
$endgroup$
Following your approach, you can combine all cases to write
$$
G=Htimes mathbb{Z}/15,
$$
where $H$ is an abelian group of order $16$. An arbitrary element of $G$ can be written as
$$
g=h+k.
$$
Then, since $G$ is abelian,
$$
sigma(g)=sigma(h)+sigma(k).
$$
Use Lagrange's theorem to study $sigma(h)$ and Fermat's little theorem or Lagrange's theorem to study $sigma(k)$.
answered Dec 8 '18 at 11:10
Michael BurrMichael Burr
26.8k23262
edited Dec 8 '18 at 19:16
answered Dec 8 '18 at 11:10
Michael BurrMichael Burr
26.8k23262
answered Dec 8 '18 at 11:10
Michael BurrMichael Burr
26.8k23262
answered Dec 8 '18 at 11:10
Michael BurrMichael Burr
26.8k23262
26.8k23262
$begingroup$
$G=Htimes mathbb{Z}_{15}$, for$gin G$ ,$g=hcdot k,sigma(g)=(hk)^{16}=h^{16}k^{16}=sigma(g)cdot sigma(k)$, $h in H$ and $H$ is of order 16 hence $sigma(h)=e$ while $k in mathbb{Z}_{15}$ hence $sigma(k)=k^{16}=k$. Therefore $sigma(g)=sigma(k)=k $. $Ker={g=(h,e_k),hin H, e_k in Z_{15}$ is the identity in$ Z_{15}}$ Hence $|Ker(sigma)|=16$. Right?
$endgroup$
– LOIS
Dec 8 '18 at 13:03
$begingroup$
Yes, that's how to finish it.
$endgroup$
– Michael Burr
Dec 8 '18 at 19:16
add a comment |
$begingroup$
$G=Htimes mathbb{Z}_{15}$, for$gin G$ ,$g=hcdot k,sigma(g)=(hk)^{16}=h^{16}k^{16}=sigma(g)cdot sigma(k)$, $h in H$ and $H$ is of order 16 hence $sigma(h)=e$ while $k in mathbb{Z}_{15}$ hence $sigma(k)=k^{16}=k$. Therefore $sigma(g)=sigma(k)=k $. $Ker={g=(h,e_k),hin H, e_k in Z_{15}$ is the identity in$ Z_{15}}$ Hence $|Ker(sigma)|=16$. Right?
$endgroup$
– LOIS
Dec 8 '18 at 13:03
$begingroup$
Yes, that's how to finish it.
$endgroup$
– Michael Burr
Dec 8 '18 at 19:16
$begingroup$
$G=Htimes mathbb{Z}_{15}$, for$gin G$ ,$g=hcdot k,sigma(g)=(hk)^{16}=h^{16}k^{16}=sigma(g)cdot sigma(k)$, $h in H$ and $H$ is of order 16 hence $sigma(h)=e$ while $k in mathbb{Z}_{15}$ hence $sigma(k)=k^{16}=k$. Therefore $sigma(g)=sigma(k)=k $. $Ker={g=(h,e_k),hin H, e_k in Z_{15}$ is the identity in$ Z_{15}}$ Hence $|Ker(sigma)|=16$. Right?
$endgroup$
– LOIS
Dec 8 '18 at 13:03
$begingroup$
$G=Htimes mathbb{Z}_{15}$, for$gin G$ ,$g=hcdot k,sigma(g)=(hk)^{16}=h^{16}k^{16}=sigma(g)cdot sigma(k)$, $h in H$ and $H$ is of order 16 hence $sigma(h)=e$ while $k in mathbb{Z}_{15}$ hence $sigma(k)=k^{16}=k$. Therefore $sigma(g)=sigma(k)=k $. $Ker={g=(h,e_k),hin H, e_k in Z_{15}$ is the identity in$ Z_{15}}$ Hence $|Ker(sigma)|=16$. Right?
$endgroup$
– LOIS
Dec 8 '18 at 13:03
$begingroup$
Yes, that's how to finish it.
$endgroup$
– Michael Burr
Dec 8 '18 at 19:16
$begingroup$
Yes, that's how to finish it.
$endgroup$
– Michael Burr
Dec 8 '18 at 19:16
add a comment |
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$begingroup$
Hint: Start by writing $G$ as a direct sum of a $2$-group and the rest.
$endgroup$
– Michael Burr
Dec 8 '18 at 11:04