Podemos determinar que existe una epimorphism $G\rightarrow H$?

Question

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

Answer 1

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

Deje $f$ be the isomorphism $$f:G\times G =\left(Q_8\times D_8\right)\times \left(Q_8\times D_8\right)\longrightarrow \left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right).$$ Ahora, vamos a $\mu$ and $\lambda$ be the epimorphisms $$\begin{eqnarray*}\mu:Q_8\times Q_8&\longrightarrow&Q_8 {\small \text{ Y }} Q_8\ \lambda:D_8 \times D_8&\longrightarrow&D_8 {\small \text{ Y }}D_8\end{eqnarray*}$$ donde $A {\small \text{ Y }} B$ denotes the central product of $A$ and $B$. Then $$\mu\times \lambda:\left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)$$ es un epimorphism. La clave está en que $D_8{\small \text{ Y }} D_8\cong Q_8{\small \text{ Y }} Q_8$, 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 $H=Q_8{\small \text{ Y }} Q_8$ y forma un isomorfismo $_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.$$ Así que, en conjunto, hemos $$\newcommand{\ra}[1]{\kern-1.5 ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5 ex} \newcommand{\ras}[1]{\kern-1.5 ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5 ex} \newcommand{\da}[1]{\bigg\downarrow\subir.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} $$ y por lo tanto una epimorphism $$f(\mu\times\lambda)(1_H\times \phi):G\times G\longrightarrow H\times H.$$ Sin embargo, $Q_8{\small\text{ Y }}Q_8$ is not a homomorphic image of $Q_8\times D_8$. Así que este es un contraejemplo.

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El apéndice.

• De crédito y agradecimiento a Peter Pecado por su ayuda con el paso crucial en esta respuesta.

• Ver Prop. 3.13 de estas notas para una prueba de que $Q_8 {\small \text{ Y }} Q_8\cong D_8 {\small \text{ Y }} D_8 \not\cong Q_8 {\small \text{ Y }} D_8$.

Answer 2

Si $G$ 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)} \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 $H_i\to G_m$ y todo lo demás a la identidad.