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Can someone check my solution to the following question please:

Prove that the alternating group $A_5$ has no subgroup of order 30.

I will be assuming the following result:

Theorem 1: Let $N$ be a subgroup of a group $G$ of index 2. Prove that $N$ is a normal subgroup.

Solution: Suppose $A_5$ has a subgroup group $N$ of order 30. The order of $A_5$ is 60 and by lagrange theorem, $|A_5|/|N|=2$. Hence by the Theorem 1, $N$ is of index 2, and so $N$ is normal. But $A_5$ is a simple group. Hence we arrive at a contradiction.

I try to do a simple proof where I don't have to look into the cycle structures of $A_5$ like the first question from here: question 1 and here: question 1, nor making use of the class equation, which Hungerford's Abstract algebra 3rd edition has not introduced at that edition of the text.

Thank you in advance

Seth
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    Yes, this argument is fine assuming you are allowed to cite the fact that $A_5$ is simple. Otherwise you can argue that the abelianization of $A_5$ is trivial which is a bit easier. – Qiaochu Yuan Jul 23 '22 at 23:04
  • @QiaochuYuan I already did that exercise since it appeared in an earlier section. I also saw this: https://math.stackexchange.com/questions/159077/how-to-prove-that-a-5-has-no-subgroup-of-order-30 I am not sure the arguments within the comments for the post seem to be a bit complicated. I admit i still am not used to having to make arguments based the techniques for showing that $A_n$ is simple for all $n\geq 5.$ Basically arguing by looking at structures based on cycle decomposition. – Seth Jul 23 '22 at 23:31

2 Answers2

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Following @Qaiochu, if you know (or better can show) that $A_5$ is perfect, then it's also a piece of cake.

That's $A_5'=A_5$.

Then if $H$ is any subgroup such that $A_5/H$ is abelian, then $H\supset A_5'=A_5$. So $H=A_5$.

calc ll
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  • so basically that with $A_5$ equals it's own commutator subgroup, meaning any elements in $A_5$ can be written as a product of transpositions, and since $A_5$ can be generated by elements in the form of 3-cycles and a commutator subgroup consist of product of two elements and their inverses in $A_5$, so equal its own commutator subgroup. Since a commutator subgroup is normal hence it must equal to $A_5$ due to the fact that $A_5$ is a simple group. Is that the correct argument to show that $A_5$ is a perfect group. I know it is a bit verbose, and I did not write it out all in math notation. – Seth Jul 24 '22 at 04:09
  • Well, if you know $A_5$ is simple, then it's automatic that the commutator is, since it's a normal subgroup, either $A_5$ or the identity subgroup. It can't be the identity though, because then the abelianization of $A_5$ would be itself. But $A_5$ is not abelian. I think that $A_5$ is perfect is probably significantly less strong a statement than that it's simple. Idk how to do that off the top of my head, though. – calc ll Jul 24 '22 at 04:36
  • ah ok ok. Thank you for the suggestion. There are a lot of new things I am learning I can use to solve this problem, but I am having trouble putting it all together coherently. – Seth Jul 24 '22 at 04:38
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As you have the simplicity of $A_5$ at hand, you can prove even more, namely that $A_5$ hasn't got subgroups of order $15$, $20$ and $30$.

In general, let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$: $$X_G:=\{H\subseteq G\mid H\le G \wedge H\ne G\}.$$ If $G$ is simple, then:

$$[G:H]!\ge|G|, \space\forall H\in X_G \tag 1$$

In fact, for $H\in X_G$, the group $G$ acts by left multiplication on the left quotient set $G/H$, and this action has trivial kernel$^\dagger$. So, $G$ embeds into $S_{[G:H]}$, whence $(1)$.

So, $A_5$ can't have subgroups of order $15$, $20$ and $30$, because $(60/k)!<60$ for $k=15, 20, 30$.

$^\dagger$For $H\in X_G$, the kernel $K:=\bigcap_{g\in G}gHg^{-1}$ is a proper normal subgroup of $G$, and hence $K=\{1\}$ for the simplicity of $G$.

  • aren't you using the generalised cayley theorem? I recently proved that in an exercise in Hungerford. The thing is, Hungerford in his undergrad text doesn't go into group action much. I think what you mean by embedding is that G maps into $S_{[G:H]}$ as an injective homomorphism. won't the mapping be written something like $f:G\rightarrow G/H$? – Seth Jul 24 '22 at 17:04
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    @fizcarraldo what are the significance of $[G:H]!\geq |G|$ and $(60/k)!<60$? – Seth Jul 24 '22 at 17:06
  • The injective homomorphism is from $G$ to $S_{[G:H]}$, so the codomain (of order $[G:H]!$) must be "bigger than" the domain (of order $|G|$). And it can't be if the subgroup $H$ has order $15$, $20$ or $30$. –  Jul 24 '22 at 17:19
  • ah because the codomain is bigger than the domain, hence the mapping is injective? – Seth Jul 24 '22 at 17:24
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    Vice versa: since the map is injective, then etc. –  Jul 24 '22 at 17:33
  • ah kk. Thank you for your clarification and your answer. – Seth Jul 24 '22 at 17:38
  • can i trouble you with two more minor points. 1) if the size of codomain is bigger than that of the domain, the the mapping is injective. How do I prove this in the case of general homomorphic mapping? I do know that in the context of linear algebra it is true. Not sure if i can just simply extend it without resorting to bases. 2) you said if $G$ is simple, then you show that the size of the codomain is bigger than the domain. But if the group is simple, and doesn't that imply automatically your homomorphic map that you used is injective with trivial kernel? – Seth Jul 25 '22 at 02:19
  • Given any proper subgroup $H$, $G$ acts on the set of the left cosets by left multiplication, and this set has size $[G:H]$. This action induces a homomorphism from $G$ to $S_{[G:H]}$. The kernel is a proper normal subgroup, and hence it is the trivial one (because $G=A_5$ is simple). Therefore the homomorphism is injective, and hence the codomain must be bigger than the domain (pigeonhole principle), i.e. $(60/k)!\ge 60$, where $k:=|H|$. Is this inequality fulfilled if $k=15, 20, 30$? Then... Be careful to bolfdface implications. –  Jul 25 '22 at 05:30