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I have a question about the following theorem that I found in some research. enter image description here

Is it possible that $E$ is the identity?

I just found this elaborated proof that might help. enter image description here

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Well, given $P$, $Q$ itself is also a $p$-group with cyclic Frattini subgroup... – anon Jul 14 '12 at 15:38
up vote 2 down vote accepted

Yes, $E$ can be the identity, as anon points out. The dihedral group with $8$ elements and the quaternion group of order $8$ are examples where this happens. Another general set of examples is when $Q$ is an extra-special group of ordr $p^{2n+1}$, $p$ any prime, and $n$ any positive integer. In the case, $Z(Q) = [Q,Q] = \Phi(Q)$ has order $p,$ and these exist for evey $p$ and $n$ (in fact, there is always more than one isomorphism type of extraspecial group of those orders). The existence of extraspecial groups of order $p^{3}$ is easy- we have seen two above when $p = 2.$ For $p$ odd, take (for example), $Q = \langle x,y : x^{p} = y^{p^{2}} = 1, x^{-1}yx = y^{1+p} \rangle$. For general $n,$ take a central product of $n$ extra-special groups of order $p^{3}$. The central product of two groups $A$ and $B$ which each have center of order $p$ may be realised as the direct product, with the "diagonal subgroup" of the two (isomorphic) centers factored out.

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This is just a slightly more general version of Jack's answer. – Geoff Robinson Jul 14 '12 at 17:07

Yes, it is possible that $E$ is the identity.

Take $P$ non-abelian of order $p^3$. In this case $U = \Phi(P) \cap \Omega_1(Z(P)) = \Omega_1(Z(P)) = Z$ and so $E=1$. $Q=P$ has $\Phi(Q) = Z(Q)$ cyclic of order $p$. Furthermore such a $P$ is not a direct product of any proper non-identity subgroups, and so $E = 1$ is not only possible, it is necessary.

This is just @anon's comment with an explicit answer.

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