Is every group of order 36 solvable?

[Chris L T521]

Here's this week's problem!

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Problem: Show that all groups of order 36 are solvable.

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[Chris L T521]

This week's problem was correctly answered by Deveno. You can find his solution below.

[sp]I shall use certain results without proof.

Firstly, if $N$ is a normal subgroup of $G$, then $G$ is solvable if and only if $N$ and $G/N$ are both solvable.

First, we look at the Sylow 3-subgroup(s) of $G$. By the Sylow theorems, there can only be 1 or 4 such subgroups (since only 1 and 4 are congruent to 1 mod 3 and divide 12).

Case 1: the Sylow 3-subgroup is normal. In this case, we take $N$ to be this subgroup, and since $G/N$ has order 4, it is abelian, thus solvable. Since $N$ has order 9 = 3*3, it is a group of order $p^2$ for $p = 3$, and thus likewise abelian (this can be shown by showing if $N$ is not cyclic, then $Z(N)$ being non-trivial (every $p$-group has a non-trivial center) must have order $p$ or $p^2$...the first possibility leads to $N/Z(N)$ cyclic, which implies $N$ abelian, the second says $N$ is already abelian).

Case 2: we have 4 Sylow 3-subgroups. Suppose that 2 of them are $H$ and $K$. Now $|H \cap K|$ must be 3, since if $|H \cap K| = 1$ we have:

$|HK| = 81 > 36$, which is impossible, since $HK \subseteq G$.

Now $|H|,|K| = 9$ so we know that $H$ and $K$ (see above) are abelian, thus $H \cap K$ is normal in $H$ and $K$. This implies that $N(H \cap K)$ contains both $H$ and $K$ so $|N(H \cap K)| =$ 18 or 36.

Case 2a: $|N(H \cap K)| = 36$. Here, we see that $H \cap K$ is normal in $G$, so we may take $N = H \cap K$, and thus $N$ is clearly solvable, being cyclic. Thus we are left with showing that $G/N$ of order 12 is solvable. We proceed as before, since $G/N$ has either 1 or 4 Sylow 3-subgroups. If it has just one, then we have a normal subgroup of order 3, and the factor group is of order 4, and these are both abelian. So, without loss of generality, we may assume $G/N$ has 4 Sylow 3-subgroups. These subgroups must have trivial intersection, giving us 8 elements of order 3. This leaves room for just one Sylow 2-subgroup (of order 4), which is consequently normal. Thus, we have a normal subgroup of order 4, which is abelian, and thus solvable, while the factor group has order 3, which is then also solvable (it is cyclic).

Case 2b: $|N(H \cap K)| = 18$. In this case, we may take $N = N(H\cap K)$ since it is of index 2 in $G$. Now, $G/N$ is cyclic of order 2 (thus solvable), so we must show in this case that a group of order 18 is solvable. In this case, we see that the order of a Sylow 3-subgroup is 9, and there can only be 1 such subgroup, since no positive integer of the form 1+3k divides 6 except 1. As this subgroup is normal with index 2, we see both it and the factor group are both abelian, which concludes our argument.

(Note: it should be possible to generalize this somewhat to show that any group of order $p^2q^2$ is solvable).[/sp]