If $G$ is a direct product of simple groups, then is every simple subgroup of $G$ isomorphic to a subgroup of some factor?

Let $G=N_1\times N_2\dots \times N_n$. Suppose that $H$ is a simple subgroup of $G$. Is $H$ isomorphic to a subgroup of $N_i$, for some $N_i$?

This is a weaker version of this question, which turned out to be false:

If $G$ is direct product of simple normal subgroups, then is every simple subgroup isomorphic to some factor

• This is the same question as math.stackexchange.com/questions/1622808 – Derek Holt Jan 22 '16 at 21:57
• @DerekHolt I thought it was different. – Jorge Fernández Hidalgo Jan 22 '16 at 22:02
• Well the other question asks whether it is possible for a simple subgroup not to be isomorphic to a subgroup of some $N_i$, which is not completely dissimilar :} – Derek Holt Jan 22 '16 at 22:28

Let $\pi_i:G\to N_i$ be the projection onto the $i$th factor. Then $\pi_i(H)\neq \{1\}$ for some $i$ and, for this $i$, we must have $\ker\pi_i=\{1\}$. Hence, $H$ is isomorphic to a subgroup of $N_i$.
This is false if you require equality. Take $G=A_6\times A_6$ and $A_5\cong H=\Delta(A_5)\leq G$, where $\Delta(A_5)=\{(\sigma,\sigma)\mid \sigma\in A_5\}$.