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Let $H$ be a proper subdirect product of two quasisimple groups $G_1$ and $G_2$. We know from the definition of quasisimple group that all the normal subgroups of $G_i$ live in $Z(G_i)$, the center of $G_i$. Then applying Goursat's lemma, we know that there exists $Z_i \leq Z(G_i)$, such that $G_1/Z_1 \cong G_2/Z_2$, and $H$ satisfies $\require{AMScd}$ \begin{CD} 1 @>>> & Z_2 @>>> & H @>>> & G_1 @>>> & 1, \end{CD} \begin{CD} 1 @>>> & Z_1 @>>> & H @>>> & G_2 @>>> & 1, \end{CD} and \begin{CD} 1 @>>> & Z_1 \times Z_2 @>>> & H @>>> & G_1/Z_1 \cong G_2/Z_2 @>>> & 1. \end{CD} My question is, since we require $G_1$ and $G_2$ to be quasisimple groups, what can we say about $H$ more than Goursat's lemma? For example, is it true that $H \cong G_1 \times Z_2$ or $H \cong G_2 \times Z_1$? In the special case where there exists a surjective homomorphism $\phi: G_1 \rightarrow G_2$, I could show that $H \cong G_1 \times Z_2$.


Edit on Aug. 16: As pointed out by @DerekHolt, it is not necessarily true that $H \cong G_1 \times Z_2$ or $H \cong G_2 \times Z_1$. Now I wonder whether this weaker version is true: $H \cong G \times Z$, where $G$ is a perfect central extension of both $G_1$ and $G_2$, and $Z$ is a simutaneous subgroup of both $Z_1$ and $Z_2$. Or even weaker: is $[H, H]$ a perfect group?

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  • $\begingroup$ It is not clear to me what you are asking. What do you mean by "all the central extensions"? The possibilities are clearly limited by $G_1$ and $G_2$. For example, if we take $G_1=G_2=A_5$, then we cannot have $H = {\rm SL}(2,5)$. But note that $H$ need not be quasisimple. $\endgroup$
    – Derek Holt
    Aug 16 at 7:04
  • $\begingroup$ @DerekHolt Thanks for asking for clarification! I have updated my question. Does this help? $\endgroup$ Aug 16 at 7:27
  • $\begingroup$ I still find "can we say more about $H$?" too vague. Could you ask something more specific? What sort of thing might you want to say about $H$? $\endgroup$
    – Derek Holt
    Aug 16 at 8:02
  • $\begingroup$ @DerekHolt Good question. I was wondering whether $H$ should always be a semi-direct of two groups, or even a direct product of two groups. Like in the example I mentioned, $H$ is a direct product if there is a surjective homomorphism from $G_1$ to $G_2$. $\endgroup$ Aug 16 at 8:16
  • $\begingroup$ We can have $G_1 \cong G_2 \cong H$, which is not a (nontrivial) direct product. I think it is either quasisimple or a direct product of a central subgroup with a quasisimple group. $\endgroup$
    – Derek Holt
    Aug 16 at 8:49
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The answer to your question "is $H \cong G_1 \times Z_2$ or $Z_1 \times G_2$" is no (i.e. not necessarily).

As an example, the Schur Multiplier of $A_6$ is cyclic of order $6$, so we take $G_1$ and $G_2$ to be $2$-fold and $3$-fold coverings of $A_6$, respectively (i.e. $2.A_6$ and $3.A_6$ in ATLAS notation).

Now if we take $Z_1 = Z(G_1)$ (of order $2$) and $Z_2 = Z(G_2)$ (of order $3$), then $H = 6.A_6$ is the full covering group of $A_6$.

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  • $\begingroup$ Thanks! I guess similar construction works for $G_1 = SU(6)/\mathbb{Z}_3$, $G_2 = SU(6)/\mathbb{Z}_2$, and $H = SU(6)$. $\endgroup$ Aug 16 at 23:23

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