Let G be a group of order 2m, where m is odd. Let H be a normal subgroup of G such that |H| = 5. Show that H is a subset of Z(G).
So far, all I know is that H is isomorphic to Z_{5} and is Abelian. I also know that G/H has order divisible by 10. So, how might I prove that all elements of H are invariant under conjugation?
@birdman The order of a conjugacy class must divide the order of \(G\). Consider the conjugacy classes of \(H\), which remain in \(H\) by normality. Since \(1 \in H\), there are \(4\) more elements to consider. All cannot be in the same class, since \(4\) \does not divide \(\#G\).
... and I can't figure out how to rule out the cases \(1, 2, 2\), \(1,1, 1, 2\) or \(1, 1, 3\).
@JordiGH So the idea is that there is only one conjugacy class? Doesn't that just mean that conjugation by G is transitive over H?
@birdman I think I figured how to finish this off.
We have that \(H = <x>\) where, writing additively, \(5x = 0\). Suppose for contradiction that \(x^g = dx\) for some \(g \in G\) and \(d \in {2, 3, 4}\), where we write conjugation exponentially, i.e. that a nonzero element of \(H\) isn't fixed pointwise by conjugation.
@birdman If you iterate conjugation by \(g\), you obtain successively \(x^{g^2} = d^2 x\), \(x^{g^3} = d^3 x\) and \(x^{g^4} = d^4 x\), which are all the other nonzero elements of \(H\). In other words, the conjugacy class of \(x\) must be of size \(4\).
But this is nonsense because the order of the conjugacy class must divide \(\#G = 2m\), which rules out \(4\).
@birdman Never mind. This argument is wrong because if \(d = 4\), then the conjugacy class isn't big enough, and the original statement is written incorrectly because the dihedral group of the pentagon has trivial centre but contains a cyclic normal subgroup of size 5.
@JordiGH Wow! Thanks so much. I’m gonna check with my book.
@birdman Thus, all elements of \(H\) are fixed pointwise by conjugation, i.e. \(H \subset Z(G)\).
By the way, I should have said that wlog, \(x\) is any nonzero element of \(H\), since any of them are a generator of \(H \cong \mathbb{Z}/5\).