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.

Proof and Step-by-step explanation:

The product is given as:

∈ G and ∈ N wher N ≤ G

and

⇒

∈ N and ∈ 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

Using normality to note

∈ N

under inversion NM is closed

and also gmn ∈ MN

so, gmn = (gm)(gn)

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

⇒ gmn ∈ MN

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

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}