# central product of two groups

• Apr 17th 2013, 09:33 AM
xixi
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 $\displaystyle D=kerf=\{(h,h^{-1})|h \in H \cap K\}$. )
• Apr 20th 2013, 08:35 AM
johng
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
• Apr 20th 2013, 11:52 AM
xixi
Re: central product of two groups
Thanks, nice and precise proof!