Question

Suppose that H and are subgroups of a group G. Prove that if H is a normal subgroup of G, then HK is a subgroup of G.

Answer 1

u have \(H,K\leq G\) and \(H\trianglelefteq G\) which means
\[\large H=g^{-1}Hg=\{g^{-1}hg:h\in H\}\qquad\forall g\in G \]
right?

Answer 2

yes that looks right! but what about the rest?

Answer 3

Suppose that \(\large h_1k_1, h_2k_2\in HK\). You need to show that \(\large (h_1k_1)(h_2k_2)^{-1}\in HK\) to conclude that \(\large HK\leq G\).

Note that \(\large (h_1k_1)(h_2k_2)^{-1} = h_1k_1k_2^{-1}h_2^{-1}\). Can you see how to use the fact that \(\large H\trianglelefteq G\) to complete this problem?

Answer 4

would I need to use the g^-1hg and somehow put that into the inverse?

Answer 5

Yes, but the key part is to determine the appropriate \(\large g\in G\) and \(\large h\in H\) to use in the \(\large g^{-1} h g\) term. My hint here is that since \(\large K\leq G\), then if \(\large k\in K\) we also have \(\large k \in G\).

Answer 6

OHHHHHHH I see! Or i think i do. instead of g^-1hg I would write k^-1hk for k in G. and then I could cancel some terms etc? right?

Answer 7

Yes, so what would k and h be in this case?

Answer 8

wouldn't it just be hk which is contained in HK?

Answer 9

I mean, in order to make \(\large k^{−1}hk\) appear in the term \(\large h_1k_1k_2^{-1}h_2^{-1}\), what should you let \(k\) and \(h\) be?

Answer 10

didn't you say eariler that h would be k^-1hk and k would be replacing g in G? I guess I'm not following what you are wanting me to see.

Answer 11

I didn't say that. The observation that needed to be made is that \(\large k_1k_2^{-1}\in G\), So take \(\large g^{-1}=k_1k_2^{-1}\) and \(\large h=h_2^{-1}\) and thus \(\large g=(k_1k_2^{-1})^{-1}= k_2k_1^{-1}\).

Hence, \(\large g^{-1} hg = (k_1k_2^{-1})h_2^{-1}(k_2k_1^{-1})\) and it now follows that
\[\large h_1k_1k_2^{-1}h_2^{-1} = \underbrace{\underbrace{h_1}_{\in H}\underbrace{(k_1k_2^{-1})h_2^{-1}(k_2k_1^{-1})}_{\in H\text{ since }H\trianglelefteq G}}_{\in H}\underbrace{k_1k_2^{-1}}_{\in K} \in HK\]

Therefore \(\large HK\leq G\).

Does this make sense?

Answer 12

yes thank you for your explanation and all of your help!