central product of two groups

How can I prove that central product of $\displaystyle D_{8}$ by $\displaystyle D_{8}$ is isomorphic to central product of $\displaystyle Q_{8}$ by$\displaystyle Q_{8}$, i.e. $\displaystyle D_{8}\circ D_{8} \cong Q_{8}\circ Q_{8}$.

(If G is a group and H,k be its subgroups, then G is central product of H by K if we have: 1)G=HK. 2)hk=kh for all h in H and all k in K, then we write G=$\displaystyle H \circ K$.

Hint: We have $\displaystyle H \circ K \cong (H \times K)/D$ where $\displaystyle D=kerf $ and $\displaystyle f:H \times K \rightarrow H \circ K$ is a homomorphism for which we have $\displaystyle f(h,k)=hk$. Then $\displaystyle D=kerf=\{(h,h^{-1})|h \in H \cap K\}$. )

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Re: central product of two groups

Hi,

My first thought was that this is easy. Just compute in the dihedral group and the quaternion group. However, the details turn out to be somewhat lengthy. Here they are:

Attachment 28044

Attachment 28045

Re: central product of two groups

Thanks, nice and precise proof!