# Why is there exactly one non-solvable group of order 180?

My homework assignment for next week has the following problem in it.

Show that there exists a non-solvable group of order 180.

The problem is easy to solve given that we have already proven that $A_5$ is not solvable. Since the product of two groups is solvable iff both groups are solvable, $A_5\times\mathbb Z/3\mathbb Z$ isn't solvable, and it's clear that its order is 180.

However, I thought it would be cool to see if there are others. I found a page on the internet which says that that aren't, but without proof. I know little group theory and I'm not used to working in groups -- last time I spent a significant amount of time on groups was during my first year. This makes me completely unable to think of a way to prove this statement, or even estimate its difficulty. Could you help me with this? I don't want to specify what kind of answer I want (a proof or a hint), because I don't know what is best for me. I would like to ask you to judge what kind of answer would be best in this case.

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I haven't followed it through to see whether it would lead anywhere, but have you tried using Sylow Theorems? –  Edward Hughes May 18 '12 at 21:47

I think it is fairly easy, though perhaps a little lenghty and annoying, to show that every group of order less than 60 is solvable, thus making $\,A_5\,$ the smallest non-solvable group, thus making every finite group of the form $\,A_5\times H\,$ non-solvable.

Now, if your university's library has access to JSTOR, you have a paper determining all the (37) groups of order 180 here http://www.jstor.org/discover/10.2307/2371201?uid=3738240&uid=2129&uid=2&uid=70&uid=4&sid=56184846333

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Every group of the form $A_5 \times H$ is non-solvable, because every subgroup of a solvable group is solvable. We know that $A_5$ is not solvable and is a subgroup of $A_5 \times H$. I don't understand how this helps to show that the non-solvable group of order $180$ is unique though.. could you explain? –  Mikko Korhonen May 18 '12 at 23:21
@m.k. If you have a group of order $180 = 2^2\times 3^2\times 5$ that is not solvable, it is either nonabelian simple, or has a nonabelian simple group as a chief factor. The only nonabelian simple group of order less than $180$ whose order divides $180$ is $A_5$, so the group must be an extension of $A_5$ by something or something by $A_5$; the something must be $C_3$. For split extensions (semidirect products) we must get a direct product. So then the only question is whether there are any nonsplit extensions of $A_5$ by $C_3$ or of $C_3$ by $A_5$. –  Arturo Magidin May 18 '12 at 23:34

Let $G$ be a non-solvable group of order $180$. If $Q$ is a Sylow 3-subgroup, it not hard to see $|N_G(Q)|=18$. Since $Q$ is abelian, $N_G(Q)$ controls fusion. Since every group of order $18$ has a quotient of size $3$, we have $[G:G']\ge 3$. Since $G$ is non-solvable, so is $G'$, and thus $G'\cong A_5$. We thus have the SES $$1\rightarrow G'\rightarrow G\rightarrow C_3\rightarrow 1.$$

This is a cyclic extension, so we can let $C_3=\langle \phi\rangle$, with $\phi\in\mathrm{Aut}(A_5)\cong S_5$, with $\phi^3\in A_5$. Thus $\phi\in A_5$, and so $\phi^3=1$. Thus the SES splits, and $G\cong A_5\rtimes\langle\phi\rangle$. Since $\phi^3=1$, we have $\phi$ acting as a 3-cycle, say $(123)$. Then $\phi\cdot(132)$ is central in $G$, and not contained in $A_5$. With $N=\langle\phi\cdot(132)\rangle$ we then have $G=A_5\times N$.

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Let $G$ be a non-solvable group of order $180$.

Let's count the Sylow 5-subgroups of $G$. There can be $1$, $6$, or $36$.

If there's $1$, it's normal (call it $P$), and $G/P$ is a (solvable!) group of order $36$, so $G$ is solvable; contradiction.

If there's $36$, then a Sylow 5-subgroup $P$ is self normalizing, and being abelian, this implies $G$ has a normal subgroup of order $36$ (Burnside's transfer theorem); again, this implies $G$ is solvable. [Note that the argument here also implies if $N_G(P)$ is abelian, then $G$ is solvable.]

So $G$ has $6$ Sylow 5-subgroups; if (again) one is called $P$, we have that $|N_G(P)|=\frac{180}{6}=30$. Since this is twice an odd number, and all groups of order $15$ are abelian, we see that $|N_G(P)/C_G(P)|\le 2$. Thus if $x\in N_G(P)$ has order $3$, we also have $x\in C_G(P)$.

The conjugation action of $G$ on the Sylow 5-subgroups maps $G$ to $S_6$, and since no element of order $3$ centralizes an element of order $5$ in $S_6$, we get that the subgroup of order $3$ in $N_G(P)$ is in the kernel of this mapping $G\rightarrow S_6$. Since already then the image has size at most $\frac{180}{3}=60$ (and $G$ is non-solvable!), this must be exactly the kernel. So $G$ has a normal subgroup of order $3$; call it $N$.

[The above is just a long-winded argument to find a normal subgroup of order $3$.]

Note that since $G$ in non-solvable, for any non-trivial normal subgroup $K\lhd G$, either $|K|= 60$ or $[G:K]=60$. In particular, consider $C_G(N)$. It contains a subgroup of order $9$ (in fact it contains all of them), as well as $N_G(P)$. So it has order at least $90$, and we conclude $C_G(N)=G$, otherwise stated as: $N$ is central.

Now suppose $G$ is perfect, i.e., $G'=G$. If we let $Q$ be a Sylow 3-subgroup (which of course contains $N$), we can form the transfer map $\phi:\ G\rightarrow Q$. Since $G$ is perfect, $\phi(G)$ is trivial. But if $N=\langle n\rangle$, then $\phi(n)=n^{[G:Q]}=n^{-1}$, because $n$ is central; contradiction.

Thus $G' < G$. We don't have $G'=N$, since $G/N$ is non-abelian. Thus $G'$ has order $60$. We can't have $N\le G'$ (it implies $G$ is solvable), so we have $G'\cap N=\lbrace 1\rbrace$. In particular, $G=N\times G'$, with $G'\cong G/N\cong A_5$. We are done.

[I should note things would go a lot quicker if we used some group cohomology, or even the Schur multiplier of $A_5$.]

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Thank you very much! It may take me a while to accept this answer, because the proof is a bit above my level. But I should be able to digest it given enough time. –  user23211 May 19 '12 at 18:42