How can it be proved that any subgroup of $A_5$ has order at most 12? This is [Herstein, Problem 2.10.15], which also gives the hint that I can assume the result of the previous problem that $A_5$ has no normal subgroups $N \ne (e),A_5$.

This problem appears in an earlier section of the text than the Sylow theorems. There is a proof given at Subgroups of $A_5$ have order at most $12$?, but it uses the Sylow theorems, and I wonder if a more elementary proof is available.

So far, I can prove the following: For $n \ge 3$, the subgroup generated by the 3-cycles is $A_n$; if a normal subgroup of $A_n$ contains even a single 3-cycle it must be all of $A_n$; $A_5$ has no normal subgroups $N \ne (e),A_5$. I showed the latter by repeatedly conjugating a given nontrivial element in $A_5$ by 3-cycles to eventually obtain elements whose product is a 3-cycle.


Let $H$ be a proper subgroup of $A_5$ with $|H| > 12$, and let $A_5$ act on the set of left cosets of $H$ by left multiplication. Then, you should be able to see that the kernel of this action must be strictly between $1$ and $A_5$.

  • $\begingroup$ Thanks, that works. Let $H < A_5$, and let $S=\{g_1 H, g_2 H, \ldots \}$ be the set of left cosets of $H$. The action $\theta$ of $A_5$ on $S$ by left multiplication has image which is nontrivial, for eg, given two distinct cosets $g_1 H, g_2H$, $\exists a=g_2 g_1^{-1}$ that maps $g_1 H$ to $g_2 H$. If $|H|=15$, say, then the homomorphism $\theta$ maps a 60 element group to a $4!=24$ element group and hence has a nontrivial kernel $K$. Hence $K>1$. Since the image of $\theta$ is nontrivial, $|K|<60$. Thus, $1<K<A_5$, a contradiction. The same proof works if $|H|=20$ or $|H|=30$. $\endgroup$ – user70056 Apr 14 '13 at 5:14
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    $\begingroup$ Yes, I see that now. If $G$ is simple of order $n$, and $H$ is a subgroup of $G$ of order $k$, by the above proof, it is necessary that $n \le (n/k)!$. This means $k$ cannot be too large. $\endgroup$ – user70056 Apr 15 '13 at 13:19
  • $\begingroup$ ok, shall try to think about this sometime. $\endgroup$ – user70056 Apr 15 '13 at 16:24
  • $\begingroup$ I will add that the kernel is $\cap xHx^{-1}$, which is proven to be the largest normal subgroup of $A_5$ contained in $H$ and $A_5$ is simple, hence trivial. Then if the subgroup order is say 15, $A_5 \cong W, W \le S_4$ by first homomorphism theorem thus contradiction $\endgroup$ – Daniel Li Nov 9 '17 at 19:07
  • $\begingroup$ I have a question,if $|H| =15$, why does the homomorphism $\theta$ map a 60 element group to a $4!=24$ element group? $\endgroup$ – davkav9 May 9 '18 at 14:36

A proof using Galois correspondence: take a rational polynomial with splitting field $U$ such that $\hbox{Gal}(U/\mathbb Q)\cong A_5$ (such a polynomial exists, e.g. $x^5+20x-16$). Since $A_5$ is simple, $U$ has to be the splitting field of the minimal polynomial (over $\mathbb Q$) of each $\alpha\in U\setminus \mathbb Q$. It follows that the degree of these minimal polynomials has to be $\geq 5$ which yields $[V:\mathbb Q]\geq 5$ for any field $\mathbb Q<V\leq U$. The last statement is equivalent to the non-existence of subgroups of $A_5$ having index $2, 3$ or $4$.

  • $\begingroup$ A very interesting approach, but I don't think this qualifies as more elementary than Sylow $\endgroup$ – Hagen von Eitzen Mar 1 '19 at 15:58

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