Let $G,H$ be finite groups. Suppose we have an epimorphism $G\times G\rightarrow H\times H$ Can we find an epimorphism $G\rightarrow H$?
Can we ascertain that there exists an epimorphism $G\rightarrow H$?
-
7I already put a bounty on thi$s$. No luck. I think it's ti$m$e to $m$ake thi$s$ a millenium problem to replace the Poincaré conjecture. – 2013-05-04
2 Answers
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 the 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 the 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 $\phi:D_8{\small \text{ Y }} D_8\longrightarrow Q_8{\small \text{ Y }} Q_8,$ 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} \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 $f(\mu\times\lambda)(1_H\times \phi):G\times G\longrightarrow H\times H.$ However, $Q_8{\small\text{ Y }}Q_8$ is not a homomorphic image of $Q_8\times D_8$. 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 $p$-groups by David A. Craven", in case the link breaks again) for a proof that $Q_8 {\small \text{ Y }} Q_8\cong D_8 {\small \text{ Y }} D_8 \not\cong Q_8 {\small \text{ Y }} D_8$.
-
2@user10676: there are three central elements of order $2$: one lives in the $Q_8$ factor, one in the $D_8$ factor, and one in neither. Quotienting by either of the first two gives a group which *still* decomposes as a direct product, and that is not true for $Q_8 {\small \text{ Y }}Q_8$. The third gives the quotient $Q_8 {\small \text{ Y }}D_8$, as you remarked. – 2013-08-29
If $G$ and $H$ are both semisimple then we can decompose into a product of irreducible groups. So let $G=G_1^{(j_1)}\times ...\times G_m^{(j_m)}$ where each $G_i^{(j_i)}$ is indecomposable and $(j_i)$ is the multiplicity of $G_i$ in $G$. We can do similar for $H$. By an extension of Schur's Lemma the only homomorphisms from $G\times G\cong G_1^{(2n_1)}\times...\times G_k^{(2n_k)} \to H\times H \cong H_1^{(2j_1)}\times..\times H_m^{(2j_m) } $take $G_i$ to $H_j$ where $G_i$ and $H_j$ are isomorphic. So if we have a surjective homomorphism from $G\times G \to H\times H$ then for each $H_i^{(j_i)}$ in there is a $G_m^{(n_m)}$ with $H_i$ isomorphic to $G_m$ and $j_m \le n_m$. So clearly we can construct a homomorphism from $G \to H$ by mapping said $H_i\to G_m$ and everything else to the identity.
-
0... hum okay.... I'm edit this post Wednesday. I've got too much homework due till then to care. – 2013-04-14