# What are the normal subgroups of a direct product?

Let $N$ be a normal subgroup of $G \times H$, and let $\pi_1: G \times H \to G$ and $\pi_2: G \times H \to H$ be the canonical projections. Then $\pi_1(N)$ is normal in $G$ and $\pi_2(N)$ is normal in $H$. What else can we say? I know that it is not true, in general, that $N \simeq \pi_1(N) \times \pi_2(N)$.

I'm particularly interested in the case where $G$ and $H$ are simple. In that case, $N \simeq \pi_1(N) \times \pi_2(N)$ except possibly in the case where $\pi_1(N) = G$ and $\pi_2(N) = H$. In that case, what do we know?

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See Goursat's lemma.

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Perfect! So in particular, if I understand correctly, if $G$ and $H$ are distinct simple groups, then it is true that for a normal subgroup $N$, $N \simeq \pi_1(N) \times \pi_2(N)$. –  Gabe Cunningham May 6 '10 at 13:00
Wow, I never knew this had a name. –  HJRW May 6 '10 at 17:32

If $G$ and $H$ are non-abelian simple groups and $N$ is a normal subgroup of $G×H$, then $N$ is equal (not just isomorphic) to $1×1$, $1×H$, $G×1$, or $G×H$. If $G$ and $H$ are simple groups but not isomorphic, then the same is true. If $G$ and $H$ are isomorphic abelian simple groups, then $G×H$ is a two-dimensional vector space over some $\Bbb Z/p\Bbb Z$, and so it has $1 + (p+1) + 1$ normal subgroups, $1$ of dimension $0$, $p+1$ of dimension $1$, and $1$ of dimension $2$. Only in the last case can it happen that $N$ is not precisely equal to (and also not isomorphic to) $π_1(N)×π_2(N)$, but of course this happens for $p-1$ of the $1$-dimensional subspaces, the ones that have non-trivial projection onto both the $x$ and $y$ axes.

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Let $N$ be normal in $G\times H$. For $n=(n_1, n_2) \in N$ and $(g, 1) \in G\times H$ follows $([n_1, g], 1) = (n_1^{-1}n_1^g, 1) = n^{-1}\cdot n^{(g, 1)} \in N$ (taking the notations used in group theory: $n^g = g^{-1}ng$ etc), i.e., $[\pi_1(N), G]\times 1 \le N$ (where $[A, B]$ denotes the subgroup generated by the commutators $[a, b]$ with $a \in A, b\in B$). Also $1\times[\pi_2(N), H] \le N$, hence $[\pi_1(N), G]\times[\pi_2(N), H] \le N$.

As $[\pi_1(N), G]$ is normal in $G$, one can easily deduce for $G, H$ simple the cases described by Jack Schmidt (and also what happens in the missing case that one of the two groups is abelian but the other one not).

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Thanks, fixed the missing case. –  Jack Schmidt May 6 '10 at 14:38

The following corollary is from Normal Subgroup Growth of Linear Groups: the $(G_2, F_4, E_8)$-Theorem, by Michael Larsen and Alex Lubotzky.

Corollary 1.4: Let $G = A \times B$ be the product of two groups. If for every (finite index) normal subgroup $M \lhd A$, $Z(A/M) = \{ 1 \}$, then every (finite index) normal subgroup $N \lhd G$ is of the form $N = (N \cap A) \times (N \cap B)$.

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You can certainly have other situations, even in the simple case. For instance if $G$ and $H$ are isomorphic, the diagonal maps onto each factor.

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See Goursat's lemma ;-) –  Kevin Buzzard May 6 '10 at 12:45
The diagonal's very rarely normal! –  HJRW May 6 '10 at 17:31