# Can we ascertain that there exists an epimorphism G→HG\rightarrow H?

Let $G,H$ be finite groups. Suppose we have an epimorphism Can we find an epimorphism $G\rightarrow H$?

Let $$G=Q_8\times D_8G=Q_8\times D_8$$, where $$Q_8Q_8$$ is the quaternion group and $$D_8D_8$$ is the dihedral group of order $$88$$.

Let $$ff$$ be an isomorphism $$f:G\times G =\left(Q_8\times D_8\right)\times \left(Q_8\times D_8\right)\longrightarrow \left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right).f:G\times G =\left(Q_8\times D_8\right)\times \left(Q_8\times D_8\right)\longrightarrow \left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right).$$
Now, let $$\mu\mu$$ and $$\lambda\lambda$$ be epimorphisms $$\begin{eqnarray*}\mu:Q_8\times Q_8&\longrightarrow&Q_8 {\small \text{ Y }} Q_8\\ \lambda:D_8 \times D_8&\longrightarrow&D_8 {\small \text{ Y }}D_8\end{eqnarray*}\begin{eqnarray*}\mu:Q_8\times Q_8&\longrightarrow&Q_8 {\small \text{ Y }} Q_8\\ \lambda:D_8 \times D_8&\longrightarrow&D_8 {\small \text{ Y }}D_8\end{eqnarray*}$$
where $$A {\small \text{ Y }} BA {\small \text{ Y }} B$$ denotes the central product of $$AA$$ and $$BB$$. Then $$\mu\times \lambda:\left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)\mu\times \lambda:\left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)$$
is an epimorphism. The key is that $$D_8{\small \text{ Y }} D_8\cong Q_8{\small \text{ Y }} Q_8D_8{\small \text{ Y }} D_8\cong Q_8{\small \text{ Y }} Q_8$$, so if we take an isomorphism $$\phi:D_8{\small \text{ Y }} D_8\longrightarrow Q_8{\small \text{ Y }} Q_8,\phi:D_8{\small \text{ Y }} D_8\longrightarrow Q_8{\small \text{ Y }} Q_8,$$ then we can take $$H=Q_8{\small \text{ Y }} Q_8H=Q_8{\small \text{ Y }} Q_8$$ and form an isomorphism
$$1_H\times \phi:\left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8 \right)=H\times H.1_H\times \phi:\left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8 \right)=H\times H.$$
So, all in all, we have
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \left(Q_8\times D_8\right) \times \left( Q_8 \times D_8 \right)& \ra{f} &\left(Q_8\times Q_8\right) \times \left( D_8 \times D_8 \right)&\\ & & \da{\mu\times \lambda} & & & & \\ & & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8\right) & \ras{1_H\times \phi} & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8\right) \end{array} \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \left(Q_8\times D_8\right) \times \left( Q_8 \times D_8 \right)& \ra{f} &\left(Q_8\times Q_8\right) \times \left( D_8 \times D_8 \right)&\\ & & \da{\mu\times \lambda} & & & & \\ & & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8\right) & \ras{1_H\times \phi} & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8\right) \end{array}$$
and thus an epimorphism $$f(\mu\times\lambda)(1_H\times \phi):G\times G\longrightarrow H\times H.f(\mu\times\lambda)(1_H\times \phi):G\times G\longrightarrow H\times H.$$
However, $$Q_8{\small\text{ Y }}Q_8Q_8{\small\text{ Y }}Q_8$$ is not a homomorphic image of $$Q_8\times D_8Q_8\times D_8$$. So this is a counterexample.

Appendix.

• Credit and thanks to Peter Sin for his help with the crucial step in this answer.

• See Prop. 3.13 of these notes (“The Theory of $$pp$$-groups by David A. Craven”, in case the link breaks again) for a proof that $$Q_8 {\small \text{ Y }} Q_8\cong D_8 {\small \text{ Y }} D_8 \not\cong Q_8 {\small \text{ Y }} D_8Q_8 {\small \text{ Y }} Q_8\cong D_8 {\small \text{ Y }} D_8 \not\cong Q_8 {\small \text{ Y }} D_8$$.