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The only nontrivial normal subgroups of $A_5 \times A_5$ are $A_5 \times 1$ and $1\times A_5$.

What are the normal subgroups of $(A_5 \times A_5) \rtimes C_2$? Is $A_5 \times A_5$ the only (nontrivial) one?

In general, if $G$ is a nonabelian simple group, I think the only normal subgroups of $G^n$ are the subgroups of the form $H_1 \times \cdots \times H_n$, where each $H_i$ is either $1$ or $G$. What are the normal subgroups of $G^n \rtimes S_n$?

1st Update: I just asked GAP and apparently there are four non-trivial normal subgroups of $(A_5 \times A_5) \rtimes C_2$. Besides the obvious one, $A_5\times A_5$, there are two isomorphic to $A_5$, and one isomorphic to $S_5$. (See GAP code below.) (wrong!)

2nd Update: As SteveD noticed, there was a bug in my GAP code, so I've deleted it. The group I had chosen for $C_2$ actually normalized the second $A_5$ factor, so the resulting group was isomorphic to $S_5\times A_5$, instead of $(A_5 \times A_5) \rtimes C_2$, which explains why I found the normal subgroups mentioned above. (SteveD provides alternative GAP code in his answer below.)

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You need $G$ nonabelian for the claim about subgroups to be true. – Arturo Magidin Feb 25 '12 at 23:29
@Arturo Thanks for catching that! – William Feb 25 '12 at 23:32
Can I ask how you asked GAP? If $N$ is a normal subgroup of $G$, then $N\cap (A_5\times A_5)$ is also a normal subgroup of $G$ (and of $A_5\times A_5$). If $N\cap (A_5\times A_5)$ is just a copy of $A_5$, then the involution in $C_2$ would not normalize it! – user641 Feb 25 '12 at 23:50
@SteveD The GAP code is above, and, unless there is a bug in my code, I think GAP is telling us the truth. I was thinking along the same lines as you, and then (after the helpful hints from GAP!) I realized one of the $A_5$ must be the diagonal subgroup of $A_5\times A_5$. That is normalized by $C_2$, right? Now, where are the other normal subgroups coming from? – William Feb 25 '12 at 23:56
The diagonal subgroup is not normalized by either copy of $A_5$!! – user641 Feb 25 '12 at 23:58
up vote 4 down vote accepted

If your action of $C_2$ is non-trivial, there are only the obvious three: trivial subgroup $\lbrace 1\rbrace$; whole group $G$; and the $A_5\times A_5$ subgroup.

To see this, let $H=A_5\times A_5$ and $K$ be a copy of the $C_2$. If $N$ is a normal subgroup of your group, then $N\cap H$ is a normal subgroup of both $H$ and $G$. Because it is normal in $H$ - and $H$ is semisimple - $N\cap H$ is either all of $H$; trivial; or one of the $A_5$ factors. If it's all of $H$, that case is easy. If it's trivial, then $N$ is a conjugate of $K$, and that can't happen. If it's just one $A_5$ factor, it won't be normalized by $K$, contradicting that $N\cap H$ is normal in $G$.

I am unclear on what GAP code you used, but the following works for me:

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Excellent! Thank you, Steve... and I see no reason why this argument shouldn't generalize to answer my second question, but I'll verify. – William Feb 26 '12 at 0:07
Well, there will be a couple more subgroups since $S_n$ is not simple, but the basic idea is the same. – user641 Feb 26 '12 at 0:14

You are correct that the only normal subgroups of $G^n$ are as you describe when $G$ is simple and nonabelian:

Proposition. If $G$ is nonabelian and simple, $n\geq 1$, and $N\triangleleft G^n$, then $N=H_1\times\cdots \times H_n$, where $H_i=\{1\}$ or $H_i=G$ for each $i$.

Proof. The projection of $N$ onto the $n$th component is normal in $G$, hence is either $\{1\}$ or all of $G$. Since $N$ is a subgroup of $\pi_1(N)\times\cdots\times \pi_n(N)$, we may assume that all projections are equal to $G$, and prove that $N=G^n$.

Let $g\in G$, and let $(g,g_2,\ldots,g_n)\in N$ be an element with first coordinate equal to $g$. For every $x\in G$ we have $$[(g,g_2,\ldots,g_n),(x,1,\ldots,1)] = ([g,x],1,\ldots,1)\in N;$$ thus, $[G,G]\times\{1\}\times\cdots\times\{1\}\subseteq N$; since $[G,G]=G$, it follows that $N$ contains $G\times\{1\}\times\cdots\times \{1\}$. Similar arguments show $N$ contain the $i$th copy of $G$ for each $i$, so $N=G\times\cdots\times G$. $\Box$

Now, for the problem you have, you want to look at $(A_5\times A_5)\rtimes C_2$. If $N$ is normal, then $N\cap (A_5\times A_5) \triangleleft A_5\times A_5$, so the intersection is either trivial, equal to $A_5\times\{1\}$, to $\{1\}\times A_5$, or $N$ contains $A_5\times A_5$. In the latter case, $N=A_5\times A_5$ or $N=(A_5\times A_5)\rtimes C_2$.

If $A_5\times\{1\}\subseteq N$, then for every $g\in A_5$ we have $$((1,1),1)((g,1),0)((1,1),1) =((1,g),1)\in N$$ hence we would have $A_5\times A_5\subseteq N$; symmetrically if $\{1\}\times A_5\subseteq N$.

So we are left with the case in which $N\cap (A_5\times A_5)=\{1\}$.

If $((x,y),1)\in N$, then $$((x^{-1},y^{-1}),0)((x,y),1)((x,y),0) = ((1,1),1)((x,y),1) = ((y,x),1)\in N$$ and therefore, $((x,y),1)((y,x),1) = ((x^2,y^2),0)\in N$. Therefore, $x^2=y^2=1$. So any nontrivial element of $N$ must have $A_5\times A_5$ coordinate of exponent $2$. But if $((x,y),1)\in N$ with $x\neq 1$, then there exists $z\in A_5$ such that $xz$ is not of order $2$; then $$((z,1),0)((x,y),1)((z^{-1},1),0) = ((zx,y),1)((z^{-1},1),0) = ((xz,yz^{-1}),1)\in N$$ which contradicts our computations above. Thus, no element of $N$ can have nontrivial $A_5\times A_5$ component. But then $N=\{1\}$ or $N=C_2$, and $C_2$ is not normal in $(A_5\times A_5)\rtimes C_2$. Thus, $N$ is trivial.

So it seems to me that the only three normal subgroups of $(A_5\times A_5)\rtimes C_2$ (with the nontrivial action) are the trivial subgroup, $A_5\times A_5$, and the whole group.

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Thank you very much for your detailed answer. I wish I could accept both yours and SteveD's answer. I'm conflicted since your answer is more detailed and probably took more time to write, while SteveD's answer was first and he also found the mistake in my GAP code. What should I do? What is the standard math.stackexchange protocol here? In any case I sincerely appreciate your help. Thank you!! – William Feb 26 '12 at 0:18
@William: You should accept whichever answer you want, using whatever metrics you wish to gage which one was "more helpful"; that includes "came first", "found mistake in GAP", etc. I assure you I do not take it amiss when other people's answers are accepted instead of mine! – Arturo Magidin Feb 26 '12 at 0:24
I will vote for Arturo's answer, because it took more effort - it looks much nicer! - and because he filled in details I had skipped over. – user641 Feb 26 '12 at 0:25
Thank you Arturo. I've decided to accept SteveD's answer since it was first, since he found the flaw in my code, and since you both provide perfectly good answers to the same question (the more specific question of the title). But again, thank you very much for your thoughtful and helpful answer. Judging by the upvotes, it appears to be helping other people as well. – William Feb 26 '12 at 0:30

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