## Every group of order 63 is not simple.

Recently I have been browsing the book “Topics in Group Theory” by Geoff Smith and Olga Tabachnikova. This is a great book to learn advanced group theory at the undergraduate level. I only wish I knew this book earlier in my studies…

The Chapter 4 of the book is titled “Entertainment”. It includes applications of the tools such as group actions and Sylow Theorems to prove a variety of results for both finite and infinite groups. For the finite case, the focus is partly on teaching methods to prove a statement of the form “Every group of order $n$ is not simple.” These are fun problems in my experience. I plan to do a blog post in the future outlining some general tricks concerning this topic. (The algebra qualifying exams tend to ask questions of this type).

For now, let us consider the groups of order $63$. Can we prove that every group of order $63$ is not simple? Yes:

Easy way: Note that the number of $7$-Sylow subgroups is divisor of $9$, and is $\equiv 1 (\textrm{mod } 7)$. Hence, there is only one $7$-Sylow subgroup, which must therefore be normal.

In their book, Smith and Tabachnikova uses different proof, which I will call the “hard way”. Of course, the authors are fully aware of the simpler proof above, and in fact they hint in parenthetical remark that “but the alert reader will spot a faster proof!” But I quite like the approach they take. It seems like a neat trick to try on other problems.

Hard (but still elegant) way: Let $G$ be a group of order $63$. Let’s consider $3$-Sylow subgroups of $9$. By Sylow Theorems, the number of $3$-Sylow subgroups is either $1$ or $7$. If it is one, then the $3$-Sylow subgroup is normal and $G$ is not simple, as claimed. So let’s assume that there are seven $3$-Sylow subgroups. Take two distinct $3$-Sylow subgroups $P$ and $Q$. Let $T=P\cap Q$. Clearly, $|P|=|Q|=9$ and so by Lagrange’s Theorem $|T|=1$ or $|T|=3$. But if $|T|=1$, then the subset $PQ=\{pq: p\in P, q\in Q\}$ would contain $|P|\cdot |Q|=81$ distinct elements which is absurd since the group has only $63$ elements in total! Thus, $|T|=3$. Now we use the fact that in a $p$-group, any proper subgroup is strictly contained in its normalizer (it is a nice exercise; the proof by induction on the order of the group works smoothly. Alternatively, we have the more general fact that in any finite group $G$, if $H$ is a $p$-group inside $G$, then $[N_{G}(H): H]\equiv [G:H] (\textrm{mod } p)$, and this is proved by considering the fixed points for the left-multiplication action of $H$ on the set of left cosets of $H$). It follows that $N_{P}(T)=P$ and $N_{Q}(T)=Q$. Therefore, $P\subset N_{G}(T)$ and $Q\subset N_{G}(T)$. Since $P$ is maximal in $G$, and $Q\subsetneq P$, it follows that $N_{G}(T)=G$ and the subgroup $T$ is normal in $G$.