If N and M are normal subgroups of G, and MN = {mn|m elementof M, n elementof N}, prove that MN is also a normal subgroup of G.

Here, consider the product of 2 elements 'MN' as [tex]n_{1} m_{1}[/tex] and [tex]n_{2}m_{2}[/tex] where [tex]n_{1} and n_{2}[/tex] ∈ N and [tex]m_{1} and m_{2}[/tex] ∈ M

The product is given as:

[tex](n_{1}m_{1})(n_{1}m_{1}) = n_{1}(n_{2}m_{1})m_{2}[/tex]

[tex]m_{1}[/tex] ∈ G and [tex]n_{2}[/tex] ∈ N wher N ≤ G

and [tex]m_{1}n_{2}m_{1}^{-1} = n_{3} n_{2}m_{1} = m_{1}n_{3}[/tex]

⇒ [tex](n_{1}m_{1})(n_{1}m_{1}) = n_{1}(n_{2}m_{1})m_{2} = n_{1}m_{2}n_{2}m_{1}[/tex]

[tex]n_{3} and n_{1}[/tex] ∈ N and [tex]m_{1} and m_{2}[/tex] ∈ M

The product is in NM, under multiplication NM is closed

Now, we will consider the inverse of element of an element of NM as

[tex](nm)^{-1} = n^{-1}m^{-1}[/tex]

Using normality to note

[tex](n)_{1} = n^{-1} m m^{-1}[/tex] ∈ N

[tex](nm)^{-1} = n^{1}m^{-1}[/tex]

under inversion NM is closed

and also gmn[tex]g^{-1}[/tex] ∈ MN

so, gmn[tex]g_{-1}[/tex] = (gm[tex]g_{-1}[/tex])(gn[tex]g_{-1}[/tex])

Here, first term belongs to M and the second term to N

⇒ gmn[tex]g_{-1}[/tex] ∈ MN

Hence proved that MN is a normal sub group of 'G'