Let $G$ be a finite group such that $G$ has a normal subgroup $H$ and $H$ is isomorphic to the alternating group $A_5$. Also we know that $G/H \cong A_5$.

Can we say that $G \cong A_5\times A_5$? Thanks for your helps

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    $\begingroup$ We can say that $G \cong A_5 \times A_5$. This is hardly a research level question. $\endgroup$
    – Derek Holt
    May 15, 2014 at 7:50

2 Answers 2


Well, $G$ must be isomorphic to the direct product $A_{5} \times A_{5}$ in any case, under your assumptions, since the outer automorphism group of $A_{5}$ just has order $2$. Note that $H$ is a maximal normal subgroup of $G,$ and that $F(G) = 1$ under your assumptions. If $G \not \cong A_{5} \times A_{5},$ then $H = F^{*}(G),$ and $G/H$ is isomorphic to a subgroup of ${\rm Out}(H)$ (it is always the case for any finite group $X$ that $X/F^{*}(X)$ is isomorphic to a subgroup of ${\rm Out}(F^{*}(X)).$

Here is a more direct proof without using prior knowledge of ${\rm Aut}(A_{5})$ : we have $G = HN_{G}(P)$ for a Sylow $5$-subgroup $P$ of $G.$ Hence $[N_{G}(P):N_{H}(P)] = [G:H] = 60.$ There is an element $x \in N_{G}(P) \backslash N_{H}(P)$ whose order is a power of $3.$ Now $x$ centralizes $P,$ since $[N_{G}(P):C_{G}(P)]$ divides $4$. Since $H$ has $6$ Sylow $5$-subgroups, and $x$ permutes the Sylow $5$-subgroups of $H$ by conjugation, and already fixes $P,$ we see that $x$ must normalize at least one more Sylow $5$-subgroup of $H,$ say $Q.$ As with $P,$ we see that $x$ must centralize $Q$ too. Then $x$ centralizes $\langle P,Q \rangle - H$ ( note that a subgroup of $H$ with more than one Sylow $5$-subgroup has at least 6, so has order at least $30,$ whereas $H$ has no subgroup of index $2$. Thus $HC_{G}(H) >H,$ so that $G = H \times C_{G}(H)$ as $G/H$ is simple.


This is essentially the same proof as Geoff's, but in different language. The action of $G$ on $H$ by conjugation gives a homomorphism $\rho\colon G\to\text{Aut}(H)$ whose kernel is the centralizer $C_G(H)$. The image $\rho(G)$ is a subgroup of $\text{Aut}(H)\cong S_5$ which contains $\rho(H)\cong A_5$, so $C_G(H)$ has order either $60$ or $30$. Since $H$ has trivial center we have $C_G(H)\cap H=1$, so that $HC_G(H)=H\times C_G(H)$. Thus the image of $C_G(H)$ under the projection $G\to G/H$ is a subgroup of $G/H$ which is isomorphic to $C_G(H)$, meaning a subgroup of $A_5$ of order either $30$ or $60$, which can only be $A_5$. Therefore $G=H\times C_G(H)\cong A_5\times A_5$.

  • $\begingroup$ I am really thankful for your answer $\endgroup$
    – BHZ
    May 15, 2014 at 9:19
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    $\begingroup$ I really don't think that it's in the spirit of the system to first answer a question, and then vote to close it, thereby preventing other answers from being added. $\endgroup$
    – Alex B.
    May 15, 2014 at 13:10
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    $\begingroup$ @Alex: Asked on meta: meta.mathoverflow.net/questions/1700/…. $\endgroup$ May 15, 2014 at 13:45

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