## When the quotient by an intersection is isomorphic to a product of quotients

Let $G$ be a group and let $M,N \leq G$ be normal such that $G = MN$. Prove that $G/(M \cap N) \cong (G/M) \times (G/N)$.

[New, short proof]

Define $\varphi : G \rightarrow G/M \times G/N$ by $g \mapsto (gM,gN)$; certainly $\varphi$ is a homomorphism. If $g \in \mathsf{ker}\ \varphi$, then $(gM,gN) = (M,N)$, so that $g \in M \cap N$. Conversely, $M \cap N$ is contained in $\mathsf{ker}\ \varphi$.

Now we argue that $\varphi$ is surjective. To see this, let $(g_1M,g_2N) \in G/M \times G/N$. Since $G = MN$, and since $M$ and $N$ are normal, we have $(g_1M,g_2N) = (m_1n_1M,m_2n_2N)$ $= (n_1m_3M,m_2n_2N)$ $= (n_1M,m_2N)$ for some $m_i \in M$ and $n_i \in N$. Let $g = m_2n_1$; now $m_2n_1 = n_1m_4$ for some $m_4 \in M$, and evidently $\varphi(g) = (g_1M,g_2N)$. So $\varphi$ is surjective. By the first isomorphism theorem, we have $G/(M \cap N) \cong G/M \times G/N$.

[Old, long proof]

Note that every $x \in G$ has the form $x = mn$ for some $m \in M$ and $n \in N$. Define a mapping $\varphi : G \rightarrow N/(M \cap N) \times M/(M \cap N)$ as follows. If $x = mn$, then $\varphi(x) = (n(M \cap N), m(M \cap N))$.

Well defined: Suppose $x = m_1n_1 = m_2n_2$. Now $m_1n_1n_2^{-1}m_2^{-1} = 1 \in M$, so that $n_1n_2^{-1} \in M$. Thus $n_1n_2^{-1} \in M \cap N$, and we have $n_1(M \cap N) = n_2(M \cap N)$. Similarly, $m_1(M \cap N) = m_2(M \cap N)$. Thus $\varphi$ is well defined.

Homomorphism: Let $m_1n_1, m_2n_2 \in G$. Note that, since $N$ is normal in $G$, we have $m_2^{-1}n_1m_2 \in N$. Note also that $(m_2^{-1}n_1m_2n_2)(n_1n_2)^{-1} = m_2^{-1}n_1m_2n_1^{-1}$. Since $M$ and $N$ are both normal in $G$, this product is in $M \cap N$. So we have $m_2^{-1}n_1m_2n_2(M \cap N) = n_1n_2(M \cap N)$. Now $\varphi(m_1n_1m_2n_2) = \varphi((m_1m_2)(m_2^{-1}n_1m_2n_2))$ $= (m_1m_2(M \cap N), m_2^{-1}n_1m_2n_2(M \cap N))$ $= (m_1m_2(M \cap N), n_1n_2(M \cap N))$ $= (m_1(M \cap N), n_1(M \cap N)) \cdot (m_2(M \cap N), n_2(M \cap N))$ $= \varphi(m_1n_1) \cdot \varphi(m_2n_2)$. So $\varphi$ is a group homomorphism.

Surjective: If $(n(M \cap N), m(M \cap N)) \in N/(M \cap N) \times M/(M \cap N)$, we have $\varphi(mn) = (n(M \cap N), n(M \cap N))$. Thus $\varphi$ is surjective.

$\mathsf{ker}\ \varphi = M \cap N$: Suppose $x = mn \in \mathsf{ker}\ \varphi$. Then $\varphi(x) = (n(M \cap N),m(M \cap N)) = (M \cap N, M \cap N)$, so that $n,m \in M \cap N$. Hence $x \in M \cap N$. Suppose $x \in M \cap N$. Now $x \in M$, so that $\varphi(x \cdot 1) = (M \cap N, x(M \cap N)) = (M \cap N, M \cap N)$. Thus $x \in \mathsf{ker}\ \varphi$.

By the First Isomorphism Theorem, we have $G/(M \cap N) \cong N/(M \cap N) \times M/(M \cap N)$. Moreover by the Second isomorphism theorem we have $G/M \cong N/(M \cap N)$ and $G/N \cong M/(M \cap N)$. Thus $G/(M \cap N) \cong (G/M) \times (G/N)$.

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### Comments

• dayana  On July 6, 2010 at 7:55 pm

buenas noches quisiera felicitarlos por la pagina y ademas les recomiendo que vean este video en el cual se demuestra el ejercicio 7 de la secion 3.3 es otra forma

• nbloomf  On July 6, 2010 at 11:07 pm

Thank you for reading, and thank you for the link to this video! His proof is much more straightforward than mine- I have a tendency to immediately reach for the First Isomorphism Theorem when defining homomorphisms from a quotient.

• Anna  On August 31, 2010 at 12:33 pm

Hi,

I think we still can make use of 1st Homom Thm and define this map:

\Phi: AxB -> (A/C) x (B/D)

by

\Phi((a,b)) = (aC,bD).

And, show that \Phi is a surjective homom and Ker \Phi = C x D.

I guess this proof is much simpler.

• nbloomf  On September 1, 2010 at 6:47 am

This works and is true, but the problem here is a tad more general- $G$ is not necessarily the direct product of $M$ and $N$.

• Raymond  On October 12, 2011 at 1:54 am

in your proof did you forget to prove its a homomorphism?

• nbloomf  On October 13, 2011 at 10:08 am

You’re right! I’ll fix it.

• Gobi Ree  On November 28, 2011 at 3:32 am

How about this? It seems more simple!
Let $f:G \to G/M \times G/N$ by $g \mapsto (gM,gN)$. It is easy to see that $f$ is a homomorphism and $\ker f=M\cap N$. To show the surjectivity, let $(g_1M,g_2N)$ be any element of $G/M \times G/N$. Then $(g_1M,g_2N)=(m_1n_1M,m_2n_2N)=(n_1m_3M,m_2N)=(n_1M,m_2N)$. If we choose $g=n_1m_2$, $(gM,gN)=(n_1m_2M,n_1m_2N)=(n_1M,m_2n_3N)=(n_1M,m_2N)$, as desired.

• nbloomf  On November 28, 2011 at 10:46 am

Wow- that’s much better!

I’ll leave the old proof up for comparison. Thanks!