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 $p^2 nmid |G|$.






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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







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    put on hold as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 yesterday


    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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      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







      share|cite|improve this question















      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






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      edited yesterday









      Tianlalu

      2,594632




      2,594632










      asked yesterday









      mathnoob

      58311




      58311




      put on hold as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 yesterday


      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.




      put on hold as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 yesterday


      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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          Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an 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 $|G|$.






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            1 Answer
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            1 Answer
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            active

            oldest

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            active

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            active

            oldest

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            up vote
            2
            down vote













            Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an 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 $|G|$.






            share|cite|improve this answer



























              up vote
              2
              down vote













              Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an 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 $|G|$.






              share|cite|improve this answer

























                up vote
                2
                down vote










                up vote
                2
                down vote









                Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an 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 $|G|$.






                share|cite|improve this answer














                Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an 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 $|G|$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday









                Santana Afton

                2,4642629




                2,4642629










                answered yesterday









                mathnoob

                58311




                58311















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