Existence a certain subgroup of a group

‎Let ‎‎$G$ ‎be a‎ ‎finite group ‎such ‎that ‎‎$G=P\rtimes Q‎‎‎‎$ ‎where ‎‎$P\in {\rm Syl}_p(G)‎‎$;‎ ‎‎$‎‎P\cong \Bbb{Z}_p\times \Bbb{Z}_p‎$ ‎‎‎‎ and ‎$Q\in {\rm Syl}_q(G)$; ‎$‎‎|Q|=q$ (‎$‎‎p, q$ ‎are primes‎)‎‎.

Can we classify groups ‎$‎‎G$ ‎‎which‎‎‎ ‎contain a‎ ‎subgroup ‎of ‎order ‎‎$‎‎pq$? ‎ ‎(We know that the alternating group ‎$A_4‎‎$ ‎has no subgroup of order 6‎).‎

Thank you for your thoughts on this!

Yes. If $q$ does not divide $p-1$, there is 1 such group, the abelian group. If $q=2$, then there are 3 such groups. Otherwise there are $3+(q-1)/2$ such groups up to isomorphism.

The gist is that $P$ is a two-dimensional $Q$-module, and a subgroup of order $pq$ corresponds to a one-dimensional submodule. Since $p \neq q$ by your Sylow assumptions, $P$ is completely reducible, so is a direct sum of two one-dimensional representations.

If $q$ does not divide $p-1$, then the only one-dimensional representation is the trivial representation, and so the only such $G$ is $P \times Q$, an abelian group.

If $q=2$, then there are two such representations, one where the element $x$ of order 2 acts as the identity, $x^{-1} y x = y$ for $y \in P_0$, and one where $x$ acts as inversion, $x^{-1} y x = y^{-1}$ for $y \in P_1$. $P$ itself is either the direct sum of two copies of $P_0$, two copies of $P_1$, or one copy of each. Three distinct groups $G$.

If $q$ divides $p-1$ and is odd, then we get $q$ distinct representations $P_0, P_1, \dots P_{q-1}$. A semi-direct product $(P_i \oplus P_j) \rtimes Q$ is isomorphic to a semi-direct product $(P_{ik} \oplus P_{jk}) \rtimes Q$ whenever $k$ is relatively prime to $k$ (all indices taken mod $q$). Hence we get three sort of special examples: $(P_0 \oplus P_0) \rtimes Q$, $(P_0 \oplus P_i) \rtimes Q$, and $(P_i \oplus P_i) \rtimes Q$ (where different nonzero $i$ all give the same group), and then $(q-1)/2$ others all of the form $(P_1 \oplus P_i) \rtimes Q$, where the $q-1$ part of the count comes from letting $i$ be any nonzero index, and the $(q-1)/2$ is from realizing $(P_1 \oplus P_i) \rtimes Q \cong (P_i \oplus P_1 ) \rtimes Q \cong (P_1 \oplus P_{1/i}) \rtimes Q$.

• Thanks a lot! Exact, nice and comprehensive answer! – sebastian Jun 2 '13 at 19:07
• I deeply respect the exposition in your answers. – Alexander Gruber Jun 3 '13 at 3:09

I take $p$ and $q$ to be different primes.

Regard $P$ as a two-dimensional vector space over $\Bbb{Z}_{p}$. You want the representation of $Q$ on $P$ (which is implicit in the semidirect product) to be reducible.

Now $\operatorname{GL}(2, p)$ has order $(p^{2} -1)(p^{2} -p) = (p-1)^{2} p (p+1)$.

If $Q$ does acts trivially on $P$ (this is always the case when $q$ does not divide $(p-1)^{2} p (p+1)$, of course there are subgroups of order $pq$.

If $Q$ does not act trivially on $P$, then $q \ne p$ does divide $(p-1)^{2} (p+1)$. Note that the subgroup $D$ of diagonal matrices has order $(p-1)^{2}$ in $\operatorname{GL}(2, p)$. If $Q$ is contained in (a conjugate of) $D$, then the action is reducible, and there are subgroups of order $pq$.

If $q$ divides $p+1$, there are reducible and irreducible representations, while if $q$ does not divide $p+1$, then all representations are reducible.

• Many thanks for your nice answer! – sebastian Jun 2 '13 at 19:04

There is the complete classification of the groups of order $p^2q$ in

W. Burnside, Theory of Groups of Finite Order, Chapt.V, sec.59.