Abelian group G is of order $240$, $sigma(g)=g^{16}in text{Aut}(G)$, then #{ker $sigma$} =?












0












$begingroup$


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







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$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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edited Dec 8 '18 at 12:15









Andrews

3901317




3901317










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














  • 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










1 Answer
1






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oldest

votes


















0












$begingroup$

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 $sigma(k)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $G=Htimes mathbb{Z}_{15}$, 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











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1 Answer
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1 Answer
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0












$begingroup$

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 $sigma(k)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $G=Htimes mathbb{Z}_{15}$, 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
















0












$begingroup$

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 $sigma(k)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $G=Htimes mathbb{Z}_{15}$, 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














0












0








0





$begingroup$

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 $sigma(k)$.






share|cite|improve this answer











$endgroup$



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 $sigma(k)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 19:16

























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


















  • $begingroup$
    $G=Htimes mathbb{Z}_{15}$, 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


















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