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

Let G,H be finite groups. Suppose we have an epimorphism G×GH×H Can we find an epimorphism GH?


Let G=Q_8\times D_8, where Q_8 is the quaternion group and D_8 is the dihedral group of order 8.

Let f 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).
Now, let \mu and \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*}
where A {\small \text{ Y }} B denotes the central product of A and B. 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)
is an epimorphism. The key is that D_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, then we can take H=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.
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}
\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)

and thus an epimorphism f(\mu\times\lambda)(1_H\times \phi):G\times G\longrightarrow H\times H.
However, Q_8{\small\text{ Y }}Q_8 is not a homomorphic image of Q_8\times D_8. So this is a counterexample.


  • 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 p-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_8.

Source : Link , Question Author : Kerry , Answer Author : Alexander Gruber

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