If $H$ is a subgroup with prime index $p$ of a finite simple group $G$, then $p$ is the maximal prime $p$...
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Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 nmid |G|$.
group-theory
edited Nov 19 at 15:52
Tianlalu
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asked Nov 19 at 15:50
mathnoob
1,137115
closed as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 Nov 20 at 1:04
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Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 nmid |G|$.
group-theory
edited Nov 19 at 15:52
Tianlalu
2,814732
asked Nov 19 at 15:50
mathnoob
1,137115
closed as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 Nov 20 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Rebellos, jgon, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
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up vote
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down vote
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Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 nmid |G|$.
group-theory
edited Nov 19 at 15:52
Tianlalu
2,814732
asked Nov 19 at 15:50
mathnoob
1,137115
Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 nmid |G|$.
group-theory
group-theory
edited Nov 19 at 15:52
Tianlalu
2,814732
asked Nov 19 at 15:50
mathnoob
1,137115
edited Nov 19 at 15:52
Tianlalu
2,814732
asked Nov 19 at 15:50
mathnoob
1,137115
edited Nov 19 at 15:52
Tianlalu
2,814732
edited Nov 19 at 15:52
Tianlalu
2,814732
edited Nov 19 at 15:52
Tianlalu
2,814732
2,814732
asked Nov 19 at 15:50
mathnoob
1,137115
asked Nov 19 at 15:50
mathnoob
1,137115
asked Nov 19 at 15:50
mathnoob
1,137115
1,137115
closed as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 Nov 20 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Rebellos, jgon, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 Nov 20 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Rebellos, jgon, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces a homomorphism $f: G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.
edited Nov 22 at 20:37
amWhy
191k27223438
answered Nov 19 at 15:50
mathnoob
1,137115
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces a homomorphism $f: G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.
edited Nov 22 at 20:37
amWhy
191k27223438
answered Nov 19 at 15:50
mathnoob
1,137115
add a comment |
up vote
3
down vote
Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces a homomorphism $f: G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.
edited Nov 22 at 20:37
amWhy
191k27223438
answered Nov 19 at 15:50
mathnoob
1,137115
add a comment |
up vote
3
down vote
up vote
3
down vote
Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces a homomorphism $f: G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.
edited Nov 22 at 20:37
amWhy
191k27223438
answered Nov 19 at 15:50
mathnoob
1,137115
Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces a homomorphism $f: G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.
edited Nov 22 at 20:37
amWhy
191k27223438
answered Nov 19 at 15:50
mathnoob
1,137115
edited Nov 22 at 20:37
amWhy
191k27223438
edited Nov 22 at 20:37
amWhy
191k27223438
edited Nov 22 at 20:37
amWhy
191k27223438
191k27223438
answered Nov 19 at 15:50
mathnoob
1,137115
answered Nov 19 at 15:50
mathnoob
1,137115
answered Nov 19 at 15:50
mathnoob
1,137115
1,137115
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