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'

answer: the answer to your question is parallel

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.