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?
Look at the subgroup $\displaystyle \left<gH\right>\leq G\slash H\Longrightarrow \left<gH\right>=K\slash K$ , by the correspondence theorem for some subgroup $\displaystyle H\leq K\leq G$ .
Assuming the group $\displaystyle G $ is finite, we get $\displaystyle 3=|\left<gH\right>|=\left|K\slash H\right|=\frac{|K|}{|H|}\longrightarrow |K|=3|H|=12$
Tonio
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