# Isomorphisms involving Product groups

1. Apr 27, 2013

### CAF123

EDIT: To moderators, I frequent this forum and the Calculus forum and I have accidently put this in Introductory Physics. Can it be moved? Sorry for inconveniences.

1. The problem statement, all variables and given/known data
1)If $G \cong H \times \mathbb{Z}_2,$ show that G contains an element a of order 2 with the property that ag = ga for all $g \in G$. Deduce (briefly!) that the dihedral group $D_{2n+1}$ (with n $\geq 1$) is not isomorphic to a product $H \times C_2$.

2. Relevant Equations

Lagrange, cyclic subgroups,

3. The attempt at a solution
The isomorphism implies a bijective homomorphism of the form $\phi: G \rightarrow H \times \mathbb{Z}_2$ with $g \mapsto (h,z),$ g in G, h in H and z in {0,1}.
This means |G| = 2|H| and so the order of G must be even, |G| = 2k, for k in naturals. The order of elements of G (or the order of any subgroups) must divide the order of the group by Lagrangeand hence all elements in G have even order. Elements are of order $g^{2\alpha}, \alpha \in \mathbb{N}$ and each creates a cyclic subgroup. (Right?) So the possible orders of alpha are 2,4,6..2α. So G does have an element of order 2. Since the subgroup generated by this element is prime, necessarily it is cyclic which in turns implies the group is abelian. This subgroup is a subset of G, so the commutavity is 'maintanied' through the group and so ag=ga.

Suppose there was an isomorphism between $D_{2n+1}$ and $H \times C_2$. This means |H| = 2n+1, n nonnegative. So the order of H is odd and the order of $C_2$ is even and so the orders are coprime. I can't see right now how to obtain a contradiction.

Many thanks.

Last edited: Apr 27, 2013
2. Apr 30, 2013

### CAF123

Does anybody have any ideas?