## Small problem with hamiltonian groups

Alright, so I am studying Hamiltonian groups in the book "Theory of Groups" by Marshall Hall (specifically chapter 12, theorem 12.5.4). I have that $c = (a,b) = a^{-1}b^{-1}ab$ obeys that $(a^i,b) = c^i$. Now, I also have that every element is finite and, in paticular, that $a$ has order $N$ and $b$ has order $M$. The claim is that $a^p$, for $p$ a prime dividing $N$, commutes with $b$. I see no way how that should be the case and I would appriciate if someone could enlighten me.

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 This is not true. Take the Cartesion product of the $8$-element quaternion group and $\mathbb Z/3$. The element $a=(i,1)$ has order $12$. Take $p=3$, and $a^3=(−i,0)$ fails to commute with $b=(j,0)$. – Will Sawin Jun 25 at 20:51 This does present with some problems, since the fact I stated is used heavily in the proof the structure of Hamiltonian groups. Is there any book where the topic shows up? – Matt Jun 25 at 21:43 Does he mean "at least one prime dividing $N$" instead of "all primes dividing $N$"? – Will Sawin Jun 26 at 0:15 Actually, I figured it out while I was going to sleep last night. Since the group is non Albenian, the centre isn't all. Then chose $a$ so that is isn't in the centre AND! so that it has the smallest possible order of all non centre elements. Then, since the order of $a^p$ is $N/p < N$ it must be the case that $a^p$ is an element of the centre. Your comment helped me see what exactly was meant be choosing $N,M$ minimally and I thank you for it. – Matt Jun 26 at 8:19 What is an Albenian group? – Igor Rivin Jun 26 at 13:32
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