# Show $HK \leq G$

Let $G$ be a group, let $H$ be a subgroup of $G$ and $K$ be a normal group of $G$,
Show $HK \leq G$, where $HK=\{hk\vert h \in H, k\in K\}$

Proof:

Since $H$ and $K$ are subgroup of $G$, $\exists e\in H,K$ , so $ee=e\in HK$.Thus, $HK$ is not empty.

Let $x,y\in HK$,$x=h_1k_1,y=h_2k_2$ and $h_1,h_2\in H, k_1,k_2\in K$, apply the 1-step subgroup text,

\begin{align} xy^{-1}= &(h_1 k_1)(h_2 k_2)^{-1}\\ =&(h_1 k_1)(k_2^{-1} h_2^{-1})\\ =& h_1ek_1k_2^{-1} h_2^{-1}\\ =&h_1h_2^{-1}h_2k_1k_2^{-1} h_2^{-1}\\=& h_1h_2^{-1}k_1k_2^{-1}\in HK\end{align}

since $h_1h_2^{-1}\in H,k_1k_2^{-1}\in K$ and $K$ is normal to $G$

Hence, $HK\leq G$

Can anyone check where I did incorrect? Thanks

Here is an alternative solution:

It suffices to prove $HK=KH$.

of course $HK=\bigcup\limits_{h\in H} hK$

and $KH=\bigcup\limits_{h\in H} Kh$

Since $K$ is normal $hK=Kh$ and therefore:

$HK=\bigcup\limits_{h\in H} hK=\bigcup\limits_{h\in H} Kh=KH$

$h k h^{-1}$ is not equal to $k$ simply because $K$ is normal in $G$. It means $hkh^{-1} \in K$ or equivalently there exists $k'\in K$ such that $hkh^{-1} = k'$. So there is a step wrong in what you have done but easily corrected.

Note: There is a theorem which states, for $H, K \lt G, \;\;$ $HK \lt G \iff HK = KH$ which is well worth looking up.

• He didn't just show closure, he used the "one-step subgroup test", which checks everything altogether: identity, inverses and closure (assuming the the set is non-empty). – ET93 May 4 '15 at 3:51
• @ET93. Misses that. Apologies. – Ishfaaq May 4 '15 at 3:54
• $h_1h_2^{-1}h_2k_1k_2^{-1}h_2^{-1}=h_1h_2^{-1}k'\in HK$? – Simple May 4 '15 at 4:16
• @Simple: Yup. Does it. – Ishfaaq May 4 '15 at 5:12

Normality of $K$ implies $HK=KH$. Then

(1) $(HK)(HK)=HK.HK=H(HK)K=HK$.

(2) $(HK)^{-1}=K^{-1}H^{-1}=KH=HK$.

So $HK$ is subgroup.