If H and K are subgroups of G, and K is normal, then HK is a subgroup of G.

[jmjlt88]

Ignoring the fact that it is redundant at times, is this proof correct? Also, is there a way to show that same result using the fact that K is closed with respect to conjugates rather than the fact that for all a in G, aK=Ka. Thank you! :)





Proposition: If H and K are subgroups of G, and K is normal, then HK is a subgroup of G.
Proof: Let H,K be subgroups of G with K normal. We wish to prove that HK is a subgroup of G.

Closure
Take any two elements of HK, say hk and h’k’. Then, we want to show that (hk)(h’k’) is in HK. We note that K is normal, which implies that for kh’, there is some k’’ in K such that kh’=h’k’’ since for all a in G, aK=Ka. Thus,
(hk)(h’k)’=h(kh’)k’=h(h’k’’)k’=hh’k’’k’. Then since hh’ is in H and k’’k’ is in K, we have that hkh’k’ is in HK.

Identity
Since H and K are both subgroups and therefore contain e, we have that HK must also contain e.

Inverses
Take any element hk in HK. We wish to show that (hk)^-1 is in HK. We note that (hk)^-1=k^-1h^-1. Then since K is normal and for any a in G, we have that aK=Ka, there exists some k’ in K such that k^-1h^-1= h^-1k’. And, we have that h^-1k’ is in HK. Hence, k^-1h^-1 is HK and every element in HK has an inverse in HK.


Therefore, we conclude that HK is a subgroup of G.

QED

 

[micromass]

That seems alright.

If you want to work with conjugates, then perhaps doing something like follows could work:

(hk)(h^\prime k^\prime)=hh^\prime [(h^\prime)^{-1} k h^\prime] k^\prime

and use that K is closed under conjugates.