Question

If N and M are normal subgroups of G, and MN = mn, prove that MN is also a normal subgroup of G.

Answer 1

Proof and 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'