## Group rambling

So today, I was trying to prove the following proposition (from the past PMATH 346 final): if $G$ is a finite $p$-group, then $G$ is solvable. Clearly if $G$ is abelian, we are done. My attempt was induction on $n$, where $|G| =p^n$.

One thing we can observe is that $Z(G)$ is nontrivial, since $G$ is a $p$-group; furthermore $|Z(G)| = p^k$ for some $k$ ($0 < k < n$). Intuitively, I thought that any group with nontrivial center would certainly satisfy $G' \neq G$ (i.e. the group is not perfect), where $G' = [G,G]$ is the commutator subgroup of $G$, and hence I could apply the inductive hypothesis to $G'$ to obtain that it is solvable, and hence $G$ would be solvable.

The implications of this, however, would be that any finite group with nontrivial center is solvable, since any nontrivial perfect group is clearly not solvable. This seems intuitively wrong, and so I’m asking myself for an example of a perfect group with nontrivial center. This isn’t so easy: we know the smallest non-solvable group (and hence the smallest perfect group) is $A_5$, the alternating group on 5 letters, which has order 60. This group is centerless, so our search doesn’t terminate here. However, if we consider the special linear group $G= SL(2,5)$, note that its center consists of the matrices $\pm I$, hence is nontrivial. On the other hand, $G$ is perfect.

The center is usually thought of as a measure of how much the commutative law holds in a group, with the best case being $Z(G) = G$ (implying $G$ is abelian). On the other hand, the commutator subgroup seems to measure how much the commutative law fails. So intuitively, it seems unlikely that a perfect group could have non-trivial center. However, this example shows us that our intuition is wrong, and the two notions are perhaps not as closely related as we think.

Another interesting theorem is the fact that if $N \unlhd G$ and both $N$ and $G/N$ are solvable, then $G$ must be solvable. It of course follows that if $G$ is a (semi)direct product of two solvable groups, then it too is solvable. However, suppose a solvable group $G$ can be realized as an internal semidirect product of two subgroups $N \unlhd G$ and $K \leq G$. Is it the case, then, that both $N$ and $K$ must be solvable?

Also, the other day, we asked the following: give an example of a group $G$ with two elements $x$ and $y$, both having finite order, such that $o(xy) = \infty$. If such a group doesn’t exist, there’s no reason why we can’t define the so-called torsion subgroup for non-abelian groups. However, all the examples of concrete groups I thought of didn’t seem to satisfy this. Jimmy gave the following nice example:

$\langle x, y : x^3 = y^3 = 1 \rangle$.