The answer to “Are two subgroups of a finite $p$-group $G$, of the same order, isomorphic?” is definitely no. Such groups are very rare. How rare?

Can you classify all finite $p$-groups $G$ such that if $H,K \leq G$ with $|H|=|K|$, then $H \cong K$?

  • 2
    $\begingroup$ I'll repeat my initial analysis here: It is easy to see that for abelian groups, only the two "extremes" satisfy this. For $p=2$ the only possible non-abelian examples are the generalized quaternion groups (and then the question is whether these satisfy it). For $p\geq 3$ any non-abelian example will have all elements of order $p$. $\endgroup$ – Tobias Kildetoft May 13 '14 at 18:03
  • $\begingroup$ trivially, all elementary-$p$ groups lie in that class. $\endgroup$ – mesel May 13 '14 at 18:03
  • $\begingroup$ @mesel You mean elementary abelian? $\endgroup$ – Tobias Kildetoft May 13 '14 at 18:04
  • $\begingroup$ [I've written up Tobias's analysis, but haven't finished $p$ odd. Hoping to leverage large abelian subgroups in one or two ways. ] $\endgroup$ – Jack Schmidt May 13 '14 at 18:04
  • $\begingroup$ math.stackexchange.com/questions/151243/… and thompson stuff are the current ideas $\endgroup$ – Jack Schmidt May 13 '14 at 18:05

Thanks to Tobias and Derek!

Theorem: If $G$ is a finite $p$-group in which every two subgroups of the same order are isomorphic then $G$ is either:

  • cyclic
  • elementary abelian
  • quaternion of order 8
  • extra-special of order $p^3$ and exponent $p$

Proof: As mentioned by Tobias, either $G$ has an elementary abelian subgroup of order $p^2$ or $G$ is cyclic, or $G$ is generalized quaternion.

In the last case, only the quaternion group of order $8$ actually works, since larger quaternion groups contain both cyclic groups of order $8$ and a quaternion group of order $8$. In the second to last case, cyclic groups just work.

Now if $G$ has an elementary abelian subgroup of order $p^2$, then $G$ has no elements of order $p^2$, lest it have a cyclic subgroup of order $p^2$. Hence $G$ has exponent $p$. In the case $p=2$, this implies $G$ is elementary abelian.

So we suppose $p$ is odd and $G$ has an elementary abelian subgroup $V$ of order $p^k$ but none of order $p^{k+1}$. Assume $G \neq V$, and let $H$ contain $V$ with index $p$. Since $H$ has exponent $p$, $H=\langle h \rangle \ltimes V$ for any $h \notin V$. Since $H$ is not elementary abelian, there must be some $v_1 \in V$ with $[h,v_1]=v_2 \neq 1$. Defining $[h,v_i] = v_{i+1}$, we get that the non-identity elements of $v_i$ are linearly independent. Suppose $v_n \neq 1$ but $v_{n+1}=1$. Note that $n \geq 2$ by definition. Then $\langle h, v_{n-1}, v_n \rangle$ is extra-special of order $p^3$. By our hypothesis, $\langle v_{n-2}, v_{n-1}, v_n \rangle$ would also be extra-special rather than abelian. Hence $|V|=p^2$.

By Derek: Now assume $G \neq H$ and let $K$ contain $H$ with index $p$. Choose some copy of $V \leq Z_2(K)$ that is normal in $K$. Then $K/V \to Z(H):x \mapsto [x,v_1]$ cannot be injective, since the domain has order $p^2$, so let $z$ be non-identity in the kernel. Then $\langle z,v_1,v_2\rangle$ is abelian, a contradiction. Hence $H=G$ or $V=G$. $\square$

  • $\begingroup$ Consider an example $G$ of order $p^4$ with centre $\langle a \rangle$ and elementary abelian normal subgroup $\langle a,b \rangle$ of order $p^2$. Since commutators with $b$ are central and powers of $a$, we have $[xy,b] = [x,b][y,b]$ for all $x,y \in G$, and since $|G/\langle a,b \rangle| = p^2$, there must exist $c \in G \setminus \langle a,b \rangle$ with $[c,b]=1$, but then $\langle a,b,c \rangle$ is elementary abelian of order $p^3$. $\endgroup$ – Derek Holt May 13 '14 at 19:31
  • $\begingroup$ That looks good, thanks! $\endgroup$ – Jack Schmidt May 13 '14 at 19:47
  • $\begingroup$ I cleaned up the part before your comment too (no need for order $p^2$, just use something in $Z_2$). $\endgroup$ – Jack Schmidt May 13 '14 at 19:52

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