Suppose that H is a normal subgroup of G. If | H|=4 and gH has order 3 in G/H, find a subgroup of order 12 in G.
How would know what the subgroup of order 12 in G would look like?
Let K=. Since |H|=4 and gH has order 3 in G/H, |K|=12. Lets now prove that K is a subgroup by the two step subgroup test.
Obviously K is a subset of H.
How would my elements look when I am doing a two step subgroup test when proving that the identity is in K, that the element ab is in K whenever a and b are in K, and a-1 is in K whenever a is in K
How K plays into my proof is rather obvious, imo. It comes from the correspondence theorem, one of the most beautiful, helpful and basic of all basic group theory theorems: Correspondence Theorem - ProofWiki
Tonio