Question

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.

Answer 1

Proof and Step-by-step explanation:

• Here, consider the product of 2 elements 'MN' as $n_{1} m_{1}$ and $n_{2}m_{2}$ where $n_{1} and n_{2}$ ∈ N and $m_{1} and m_{2}$ ∈ M

The product is given as:

$(n_{1}m_{1})(n_{1}m_{1}) = n_{1}(n_{2}m_{1})m_{2}$

• Now, we know that

$m_{1}$ ∈ G and $n_{2}$ ∈ N wher N ≤ G

and $m_{1}n_{2}m_{1}^{-1} = n_{3} n_{2}m_{1} = m_{1}n_{3}$

⇒ $(n_{1}m_{1})(n_{1}m_{1}) = n_{1}(n_{2}m_{1})m_{2} = n_{1}m_{2}n_{2}m_{1}$

• Now, we know that

$n_{3} and n_{1}$ ∈ N and $m_{1} and m_{2}$ ∈ 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

$(nm)^{-1} = n^{-1}m^{-1}$

Using normality to note

$(n)_{1} = n^{-1} m m^{-1}$ ∈ N

$(nm)^{-1} = n^{1}m^{-1}$

under inversion NM is closed

• Now, to show NM is normal in G, for any g ∈ G, mn ∈ MN

and also gmn$g^{-1}$ ∈ MN

so, gmn$g_{-1}$ = (gm$g_{-1}$)(gn$g_{-1}$)

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

⇒ gmn$g_{-1}$ ∈ MN

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