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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  • 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


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

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


    \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!

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