Finding number of subgroups of order 3 in an abelian group of order 24
$begingroup$
Problem
Find the number of subgroup of order 3 in an abelian group of order 24?
Attempt
Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem
$n_3|8$ and $n_3 equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8.
Is this correct?
group-theory sylow-theory
asked Jan 16 at 14:48
blue boyblue boy
1,243613
$endgroup$
add a comment |
$begingroup$
Problem
Find the number of subgroup of order 3 in an abelian group of order 24?
Attempt
Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem
$n_3|8$ and $n_3 equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8.
Is this correct?
group-theory sylow-theory
asked Jan 16 at 14:48
blue boyblue boy
1,243613
$endgroup$
add a comment |
$begingroup$
Problem
Find the number of subgroup of order 3 in an abelian group of order 24?
Attempt
Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem
$n_3|8$ and $n_3 equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8.
Is this correct?
group-theory sylow-theory
asked Jan 16 at 14:48
blue boyblue boy
1,243613
$endgroup$
Problem
Find the number of subgroup of order 3 in an abelian group of order 24?
Attempt
Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem
$n_3|8$ and $n_3 equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8.
Is this correct?
group-theory sylow-theory
group-theory sylow-theory
asked Jan 16 at 14:48
blue boyblue boy
1,243613
asked Jan 16 at 14:48
blue boyblue boy
1,243613
asked Jan 16 at 14:48
blue boyblue boy
1,243613
asked Jan 16 at 14:48
blue boyblue boy
1,243613
asked Jan 16 at 14:48
blue boyblue boy
1,243613
1,243613
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.
Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.
answered Jan 16 at 20:19
Sam HughesSam Hughes
741114
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.
Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.
answered Jan 16 at 20:19
Sam HughesSam Hughes
741114
$endgroup$
add a comment |
$begingroup$
Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.
Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.
answered Jan 16 at 20:19
Sam HughesSam Hughes
741114
$endgroup$
add a comment |
$begingroup$
Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.
Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.
answered Jan 16 at 20:19
Sam HughesSam Hughes
741114
$endgroup$
Let $G$ be an abelian group with $|G|=24=2^3times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $nequiv 1mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are congruent to $1mod 3$ so we can rule those out.
Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.
answered Jan 16 at 20:19
Sam HughesSam Hughes
741114
answered Jan 16 at 20:19
Sam HughesSam Hughes
741114
answered Jan 16 at 20:19
Sam HughesSam Hughes
741114
answered Jan 16 at 20:19
Sam HughesSam Hughes
741114
741114
add a comment |
add a comment |
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