The Second Isomorphism Theorem states : If G is a group, N⊴G and H≤G, we know that:
N∩H⊴H and H/(N∩H)≅HN/N.
and it seems to me as if HN/N should equal H/N. Because, hnN is the same as hN. I just don't understand what's wrong with my train of thought, as surely if I were correct, the book wouldn't be using the notation HN/N.
I hope this doesn't sound like a silly question, it's really doing my head in. Thanks for the help guys!
I dont understand the answer given to the original post which I've copied above which is
"In the usual framework H and N are subgroups of a larger group G. It is usual to assume only that H is a subgroup and N is a normal subgroup of G. In particular, N may not be a subgroup of H (let alone normal), so H/N doesn't make sense in all cases."
my problem with this answer is I cant understand why doesn't H/N make sense if N isn't a subgroup of H, since N is a normal subgroup of G and H is a subgroup of G then for every g in G there exists gng^{-1} ∈ N , which entails hnh^{-1} ∈ N which makes N "normal" to H, such that for each coset hN = Nh and thus the operator of multiplying cosets of hN is closed, and for each hN there exists h^{-1}N such that
hN * h^{-1}N = N
which makes it so that every elemnt of H/N has an inverse, so why doesnt it make sense to look at H/N as a quotient group ?
So just to reiterate my two questions are: 1. If G is a group, N⊴G and H≤G, the 2nd theroem of isomorphism says: N∩H⊴H and H/(N∩H)≅HN/N. why isnt HN/N rewritten as H/N
- If G is a group, N⊴G and H≤G and H doesnt contain N. Why is H/N not a quotient group ?
thanks
Cay (https://math.stackexchange.com/users/250263/cay), Second Isomorphism Theorem Notation Query, URL (version: 2015-07-08): Second Isomorphism Theorem Notation Query