# group of order 36 is solvable

I need to prove that group G of order 36 is solvable. This is what I came with so far:

There is either one 3-sylow subgroup or 4 3-sylow subgroups.

-first case: if there is only one 3-sylow subgroup than P- a 3-sylow subgroup of G is normal. it is also abelian and therefor solvable. [G:P] = 4 and therefor G/P is abelian so it is solvable and we are done.

-second case: if there are four 3-sylow subgroups it means by sylow theory that there is a subgroup N of index 4 so G/N is abelian and therefor solvable and N is of order 9 so it's also abelian and solvable.

the only thing I'm missing is that in the second case I don't know how to prove that N is also normal in G.

• $N$ being normal is inconsistent with the assumption that there are four Sylow subgroups of order $9$ (it's an immediate consequence of Sylow 2). – user228113 Sep 18 '17 at 11:12

In the second case, the trick is as follows: Let $$S$$ denote the set of all 3-Sylow subgroups, so that $$|S|=4$$. Let $$G$$ act on $$S$$ by conjugation (note that this is a well-defined action). This gives a homomorphism $$\varphi : G\to S_4$$. Since $$|G|>|S_4|$$, this map cannot be injective.

So let $$N$$ be the kernel of this homomorphism. Then $$N$$ is normal in $$G$$ and $$G/N$$ is isomorphic to a subgroup of $$S_4$$. Hence, $$|N|\geq 3$$, so I claim that both $$N$$ and $$G/N$$ are solvable. Let us consider the cases:

1. $$|N| = 3$$, so $$|G/N| = 12$$. Then $$N$$ is cyclic and hence solvable. I claim that $$G/N$$ is also solvable: Consider the number $$n_3$$ of $$3$$-Sylow subgroups in $$G/N$$. If $$n_3=1$$, then we are done, but if $$n_3=4$$, then all the 3-Sylow subgroups intersect trivially, so there must be exactly $$8=4\times 2$$ elements of order 3 in $$G/N$$. Hence, all the remaining 4 elements must be in a single 2-Sylow subgroup, which must be consequently normal. Hence, $$G/N$$ is solvable.

2. $$|N|=6$$, then $$|G/N|=6$$. Once again, one of the Sylow subgroups must be normal, so both groups are solvable.

3. $$|N|=9$$, then $$|G/N|=4$$, so both are abelian - hence solvable.

4. $$|N|=12$$, then $$|G/N| = 3$$. Again apply the argument from $$1$$ to see that both are solvable.

In all cases, both $$N$$ and $$G/N$$ are solvable, so $$G$$ is solvable.

• why does it follow that N and G/N are solvable? – user7080065 Sep 18 '17 at 11:47
• @user7080065: Have added the argument. – Prahlad Vaidyanathan Sep 18 '17 at 11:57
• Since all Sylow $3$-subgroups are conjugate, the image of $\phi$ is a transitive subgroup of $S_4$, so $|G/N|$ is divisible by $4$. In fact no Sylow $3$-subgroup can normalize a different Sylow $3$-subgroup, so $|N|$ cannot be divisible by $9$, and hence $|G/N|$ is divisible by $12$. So in fact $|N|=3$ and $|G/N|=12$ is the only case that can occur. – Derek Holt Sep 18 '17 at 12:13
• More or less by definition the homomorphism $\phi$ its kernel must be contained in all the Sylow $3$-subgroups. Therefore $|N|=3$ or $|N|=9$. In the latter case $N$ would be a normal Sylow subgroups, so we can rule that out. Ergo, $|N|=3$ is the only case you need to consider. – Jyrki Lahtonen Sep 21 '17 at 15:29