When are all normal subgroups of a direct product of finite groups a direct product of normal subgroups?

Question

Let ${G_1}$ and ${G_2}$ be finite groups. When are all normal subgroups of their direct product ${{G_1}\times{G_2}}$ the direct product of normal subgroups in ${G_1}$ and ${G_2}$? So when is it true that every normal subgroup of their direct product can be represented as ${{N_1}\times{N_2}}$ where ${{N_1}\lhd{G_1}}$ and ${{N_2}\lhd{G_2}}$. I know this is true when ${G_1}$ and ${G_2}$ are finite groups of coprime order. But generally this isn't the case. For example $\langle (1,1) \rangle$ is a normal subgroup of $\mathbb{Z_2}\times\mathbb{Z_2}$. However it cannot be represented in such a way. But for some reason all normal subgroups of, for example, ${{A_4}\times\mathbb{Z_3}}$ can be represented like that. So I'm interested whether there is some sort of a criterion.

Answer 1

Edit. Okay, found a simpler-to-state condition; I'm keeping the original response below, between the two horizontal lines so that (i) it gives context; and (ii) so the folks who upvoted don't feel like they voted for something completely different. The result is:

Theorem 1. Let $G_1$ and $G_2$ be groups. The following are equivalent:

1. Every normal subgroup $M\triangleleft G_1\times G_2$ is equal to $N_1\times N_2$ with $N_1\triangleleft G_1$ and $N_2\triangleleft G_2$.
2. If $N_1\triangleleft G$ and $N_2\triangleleft G_2$, the abelian groups $N_1/[N_1,G_1]$ and $N_2/[N_2,G_2]$ are torsion and for all $x\in N_1$, $y\in N_2$ the orders of $x[N_1,G_1]$ and $y[N_2,G_2]$ are coprime.

Note that your claim about $A_4\times\mathbb{Z}_3$ would then be incorrect. Example at the bottom. You may have meant $A_4\times\mathbb{Z}_2$, which does have the property.

When the groups are finite, we get:

Corollary. Let $G_1$ and $G_2$ be finite groups. Then every normal subgroup of $G_1\times G_2$ is of the form $N_1\times N_2$ with $N_i\triangleleft G_i$ if and only if for any $N_i\triangleleft G_i$, the abelian groups $N_1/[N_1,G_1]$ and $N_2/[N_2,G_2]$ are of coprime order.

What is between the two horizontal lines was the original reply.

---

Original answer, now simplified.

This may not be the kind of "criterion" you want (it seems too unwieldly and can probably be simplified), but here is one.

Theorem. Let $G_1$ and $G_2$ be groups. The following are equivalent:

1. Every normal subgroup $M\triangleleft G_1\times G_2$ is of the form $M=N_1\times N_2$ with $N_i\triangleleft G_i$.
2. If $N_i\triangleleft G_i$, and $N_1/K_1\cong N_2/K_2$ for some subgroups $K_i\triangleleft N_i$ such that $[N_i,G_i]\leq K_i$ for $i=1,2$, then $K_i=N_i$ for $i=1,2$.

Let $\pi_i\colon G_1\times G_2\to G_i$ be the projection onto the $i$th factor.

Recall that a subgroup $H\leq A\times B$ is a subdirect product of $A$ and $B$ if and only if $\pi(H)=A$ and $\pi(B)=B$.

Lemma 1. Let $H\leq G_1\times G_2$. Then $H=H_1\times H_2$ for some subgroups $H_i$ of $G_i$ if and only if $H=\pi_1(H)\times\pi_2(H)$.

Lemma 2. Let $M\triangleleft G_1\times G_2$. Then $M$ is a normal subdirect product of $N_1\times N_2$ for subgroups $N_1\triangleleft G_1$ and $N_2\triangleleft G_2$.

Proof. Set $N_1=\pi_1(M)$ and $N_2=\pi_2(M)$. $\Box$

From Goursat's Lemma, subdirect products of $N_1\times N_2$ correspond to the following:

1. A normal subgroup $K_1\triangleleft N_1$;
2. A normal subgroup $K_2\triangleleft N_2$;
3. An isomorphism $f\colon N_1/K_1\to N_2/K_2$.

Then the subgroup is given precisely by all pairs $(a,b)$ where $f(aK_1)=bK_2$.

Below, given subgroups $A$ and $B$, $[A,B]$ is the subgroup generated by all elements of the form $aba^{-1}b^{-1}$ with $a\in A$ and $b\in B$.

Lemma 3. Let $M\triangleleft G_1\times G_2$, and let $N_i=\pi_i(M)$ for $i=1,2$. If $K_1,K_2$ are the subgroups corresponding to $M$ by Goursat's Lemma, then $[N_i,G_i]\leq K_i$ for $i=1,2$.

Proof. Let $f\colon N_1/K_1\to N_2/K_2$ be the isomorphism corresponding to $M$, so that $(a,b)\in M\iff f(aK_1)=bK_2$.

To show that $[N_1,G_1]\leq K_1$ is enough to show that for all $a\in N_1$, $x\in G_1$, we have $a^{-1}xax^{-1}\in K_1$. Since $a\in N_1$, there exists $b\in N_2$ such that $(a,b)\in M$. Therefore, $f(aK_1)=bK_2$. On the other hand, since $M\triangleleft G_1\times G_2$, then $$(x,1)(a,b)(x,1)^{-1} = (xax^{-1},b)\in M.$$ Therefore, $f(xax^{-1}K_1) = bK_2 = f(aK_1)$. Since $f$ is an isomorphism, it follows that $xax^{-1}K_1 = aK_1$, so $a^{-1}xax^{-1}\in K_1$, as desired.

A symmetric argument shows that $[N_2,G_2]\leq K_2$. $\Box$

Proof of the Theorem. We first prove $1\implies 2$ by contrapositive. Suppose 2 does not hold. Then we have $N_1\triangleleft G_1$, $N_2\triangleleft G_2$, $K_1\triangleleft N_1$ with $[N_1,G_1]\leq K_1$, $K_2\triangleleft G_2$ with $[N_2,G_2]\leq K_2$, an isomorphism $f\colon N_1/K_1 \to N_2/K_2$, and at least one of $K_1\neq N_1$ and $K_2\neq N_2$ holds. Without loss of generality assume $K_1\neq N_1$.

Define $M\leq N_1\times N_2\leq G_1\times G_2$ by $$ M= \{ (a,b)\in N_1\times N_2)\mid f(aK_1)=bK_2\}.$$ It is easy to verify that this is a subgroup of $G_1\times G_2$.

To see that it is normal in $G_1\times G_2$, let $(x,y)\in G_1\times G_2$ and $(a,b)\in M$. Then $a^{-1}xax^{-1}\in [N_1,G_1]\leq K_1$, so $xax^{-1}K_1 = aK_1$; symmetrically, $b^{-1}yby^{-1}\in K_2$, so $yby^{-1}K_2 = bK_2$. Therefore, $f(xax^{-1}K_1) = f(aK_1) = bK_2 = yby^{-1}K_2$, and by the definition of $M$, we have $$(x,y)(a,b)(x,y)^{-1} = (xax^{-1},yby^{-1}) \in M.$$ Thus, $M\triangleleft G_1\times G_2$.

Finally, let $a\in N_1-K_1$. Then $(a,e)\in N_1\times N_2$, but $f(aK_1)\neq K_2$ (because $f$ is one-to-one), so $(a,e)\notin M$. Since $N_1=\pi_1(M)$ and $N_2=\pi_2(M)$, it follows that $M\subsetneq \pi_1(M)\times \pi_2(M)$, and hence by Lemma 1 we have that $M$ is not the product of a normal subgroup of $G_1$ and a normal subgroup of $G_2$. Thus, if 2 does not hold, then 1 does not hold.

We prove $2\implies 1$: Assume that 2 holds, and let $M\triangleleft G_1\times G_2$. Let $N_1=\pi_1(M)$ and $N_2=\pi_2(M)$. Then $M$ is a subdirect product of $N_1\times N_2$, so by Goursat's Lemma there exist $K_1\triangleleft N_1$, $K_2\triangleleft N_2$, and an isomorphism $f\colon N_1/K_1\to N_2/K_2$ such that $(a,b)\in M\iff f(aK_1)=bK_2$. By Lemma 3, we have $[N_1,G_1]\leq K_1$ and $[N_2,G_2]\leq K_2$. By $2$, this implies that $K_1=N_1$ and $K_2=N_2$. But that means that for all $a\in N_1$ and $b\in N_2$, $f(aK_1) = f(eN_1) = eN_2 = bN_2$, so $(a,b)\in M$. Thus,, $N_1\times N_2\subseteq M$, proving that we have $M=N_1\times N_2$, as desired. $\Box$

Corollary. If $G_1$ and $G_2$ are groups such that every normal subgroup $M\triangleleft G_1\times G_2$ is of the form $N_1\times N_2$ with $N_i\triangleleft G_i$, then the only quotient of $G_1$ that is isomorphic to a quotient of $G_2$ is the trivial quotient $G_1/G_1\cong G_2/G_2$. In particular, this occurs if both $G_1$ and $G_2$ are finite and $\gcd(|G_1|,|G_2|)=1$.

Proof of Corollary. Taking $N_i=G_i$, for any $K_i\triangleleft G_i$ we have $[G_i,K_i]\leq K_i$, so from the theorem we conclude that if $G_1/K_1\cong G_2/K_2$, then $K_1=G_1$ and $K_2=G_2$. $\Box$

---

New material

Okay, now let's simplify that and prove Theorem 1. I will use the following result:

Theorem 3. Let $G_1$ and $G_2$ be groups. Then every subgroup of $G_1\times G_2$ is equal to $H_1\times H_2$, with $H_1\leq G_1$ and $H_2\leq G_2$, if and only if $G_1$ and $G_2$ are torsion groups, and for every $x\in G_1$ and $y\in G_2$, $\gcd(|x|,|y|)=1$.

Proof. A proof can be found here. $\Box$

Proof of Theorem 1. Assume $2$ holds, and let $M\triangleleft G_1\times G_2$. Let $N_1=\pi_1(M)$, $N_2=\pi_2(M)$. Let $K_1=\{a\in N_1\mid (a,1)\in M\}$ and $K_2=\{b\in N_2\mid (1,b)\in M\}$. Then $K_1\times K_2\triangleleft G_1\times G_2$; and by Lemma 3, $[N_i,G_i]\leq K_i$. Moding out by $K_1\times K_2$ we may assume that each $N_i$ is central in $G_i$ and $M$ is a subgroup of $N_1\times N_2$. From 2 and Theorem 3 we have that every subgroup of $N_1\times N_2$ is of the form $H_1\times H_2$ with $H_i\leq N_i$; and since $\pi_i(M)=N_i$, it follows that $M=N_1\times N_2$, proving $1$.

Conversely, assume $2$ does not hold. Then there exist normal subgroups $N_1$ and $N_2$ of $G$ such that $\frac{N_1}{[N_1,G_1]} \times \frac{N_2}{[N_2,G_2]}$ has a subgroup $H$ that is not equal to a product of a subgroup of the first factor by a subgroup of the second factor. Let $M\leq G_1\times G_2$ be the subgroup of $N_1\times N_2$ (hence of $G_1\times G_2$) that contains $[N_1,G_1]\times [N_2,G_2]$ corresponding to $H$.

Note that if $M=M_1\times M_2$ for some (necessarily normal) subgroups $M_i\triangleleft N_i$, then $H=\frac{M_1}{[N_1,G_1]}\times \frac{M_2}{[N_2,G_2]}$, contradicting our choice of $H$. So to prove that $1$ does not hold, it is enough to show that $M$ is normal in $G_1\times G_2$.

Indeed: let $(a,b)\in M$, and $(x,y)\in G_1\times G_2$. Then $xax^{-1}[N_1,G_1] = a[N_1,G_1]$ and $yby^{-1}[N_2,G_2]=b[N_2,G_2]$, so the image of $(xax^{-1},yby^{-1})$ in $\frac{N_1}{[N_1,G_1]}\times \frac{N_2}{[N_2,G_2]}$ is the same as the image of $(a,b)$, and hence lies in $H$. Therefore, $(x,y)(a,b)(x,y)^{-1}\in M$. Thus, $M\triangleleft G_1\times G_2$, and this proves the theorem. $\Box$

---

Your claim about $A_4\times\mathbb{Z}_3$ would then be incorrect, since we can take $N_1=A_4$, $N_2=\mathbb{Z}_3$; with $[N_1,A_4]=K$ the Klein $4$-subgroup of $A_4$, the quotient $N_1/K$ is cyclic of order $3$, as is $N_2/[N_2,\mathbb{Z}_3]$. That leads, following the theorem, to the subgroup $M\leq A_4\times\mathbb{Z}_3$ consisting of the following elements: $$M = \bigl(K\times\{1\}\bigr)\cup\bigl( (123)K\times\{x\}\bigr)\cup \bigl((132)K\times\{x^2\}\bigr).$$ This subgroup is not of the form $N_1\times N_2$, since it does not contain $\bigl( (123),x^2\bigr)$, even though it contains elements that have $(123)$ in the first coordinate and elements with $x^2$ in the second coordinate. It is a subgroup because it corresponds to the isomorphism $f\colon A_4/K\to\mathbb{Z}_3$ given by $f\bigl((123)K\bigr) = x$. And it is normal because the conjugacy class of $(123)$ in $A_4$ is precisely the coset $(123)K$, and likewise the conjugacy class of $(132)$ in $A_4$ is precisely $(132)K$.

On the other hand, if we replace $\mathbb{Z}_3$ with $\mathbb{Z}_2$, then the group does have the desired property: because if $N_1=A_4$, then $N_1/[N_1,A_4]$ has order $3$; and if $N_1=K$ then $N_1/[N_1,A_4]$ is trivial. So the quotients $N_1/[N_1,A_4]$ and $N_2/[N_2,\mathbb{Z}_3] \cong N_2$ will have relatively prime order.

This example also shows we cannot replace $N_i/[N_i,G_i]$ with the abelianization of $N_i$ in Theorem 1, since $K\leq A_4$ is already abelian and of order $4$, and so does not have coprime order to $\mathbb{Z}_2$, even though the group has the condition on normal subgroups.