Let $H,K\leq G$ and $(H,K)$ be the subgroup generated by $\{hkh^{-1}k^{-1}\,|\,h\in H,k\in K\}$. Show that $(H,K)\triangleleft H\vee K$ I working on this Exercise from Algebra by Hungerford (Exercise II.5.3(a)). It states

If $H$ and $K$ are subgroups of a group $G$, let $(H,K)$ be the subgroup of $G$ generated by the elements $\{hkh^{-1}k^{-1}\,|\,h\in H,k\in K\}$. Show that $(H,K)\triangleleft H\vee K$.

My attempt so far:
Let $h_1k_1h_1^{-1}k_1^{-1}\cdots h_nk_nh_n^{-1}k^{-1}_n$ and $h'_1k_1'\cdots h_m'k_m'$ be arbitrary elements of $(H,K)$ and $H\vee K$, respectively. Now
\begin{multline*}
(h'_1k_1'\cdots h_m'k_m')\big(h_1k_1h_1^{-1}k_1^{-1}\cdots h_nk_nh_n^{-1}k^{-1}_n\big)(h'_1k_1'\cdots h_m'k_m')^{-1}.
\end{multline*}
From here I can't figure out how to get this into a product of $[h,k]$.
I realize how disgusting the problem is getting and realized there's probably an easier way. Any help is greatly appreciated. Thanks
Edit: would this argument be okay? That if $f$ is an automorphism of $H\vee K$, then $f\big((H,K)\big)\leq (H,K)$ and if we choose $f$ to be conjugation then we are done
 A: I've figured it out:
Let $[h,k]\in (H,K)$  be arbitrary. Since $H\vee K$ is the subgroup of $G$ generated by $H$ and $K$, it suffices to show that $\tilde{h}[h,k]\tilde{h}^{-1},\tilde{k}[h,k]\tilde{k}^{-1}\in (H,K)$ for arbitrarily chosen $\tilde{h}\in H,\tilde{k}\in K$: By the fact that for any $a,b,c\in G$, $[ab,c]=a[b,c]a^{-1}[a,c]$,
\begin{align*}
\tilde{h}[h,k]\tilde{h}^{-1}&=[\tilde{h}h,k][\tilde{h},k]^{-1}\in (H,K)
\end{align*}
and (using the fact that $[a,b]=[b,a]^{-1}$
\begin{align*}
\tilde{k}[h,k]\tilde{k}^{-1}&=\left(\tilde{k}[h,k]^{-1}\tilde{k}^{-1}\right)^{-1}=\left(\tilde{k}[k,h]\tilde{k}^{-1}\right)^{-1}\\
&=\left([\tilde{k}k,h][\tilde{k},h]^{-1}\right)^{-1}=[\tilde{k},h][\tilde{k}k,h]^{-1}\\
&=[h,\tilde{k}]^{-1}[h,\tilde{k}k]\in (H,K).
\end{align*}
A: Recall that $ H\vee K $ is the join of $ H $ and $ K $ meaning the subgroup $ \langle H\cup K\rangle $ in $ G $. 
Let $ [h, k]=hkh^{-1}k^{-1}\in (H, K) $, it suffices to check $ \forall x\in H\vee K $ and $ \forall [h, k]\in (H, K) $, $ x[h, k]x^{-1}\in (H, K) $. 
Notice that $ x[h, k]x^{-1}=[h, k]([k, h], x) $ so it suffices to show that $ ([k, h], x)\in (H, K) $. 
Since $ x\in\langle H\cup K \rangle $ takes the form $ x=h_1k_1h_2k_2\cdots h_m k_m $ for some $ m\in\mathbb{Z}_{\geq 1} $ as $ h, k $ appear alternatively in odd and even positions. By doing the trick: $ hkh'=hkh^{-1}k^{-1}khh'=[h, k]k(hh') $ we can always write $ x=\lambda k_0 h_0 $ where $ \lambda\in (H, K) $ and $ k_0\in K, h_0\in H $. 
Hence $ ([k, h], x)=[k, h]\lambda h_0k_0 [h, k]k_0^{-1}h_0^{-1}\lambda^{-1} $ and if we can show that $ h_0k_0[h, k]k_0^{-1}h_0^{-1}\in (H, K) $ then we are done. 
Expand the item and we get $ h_0k_0hkh^{-1}k^{-1}k_0^{-1}h_0^{-1} $. 
First, $ h_0k_0hk=[h_0, k_0][k_0, h_0h][h_0h, k_0k]k_0kh_0h $; 
then we have 
        \begin{align*}
  h_0k_0hkh^{-1}k^{-1}k_0^{-1}h_0^{-1}&=[h_0, k_0][k_0, h_0h][h_0h, k_0k]k_0kh_0hh^{-1}k^{-1}k_0^{-1}h_0^{-1}\\
  &=[h_0, k_0][k_0, h_0h][h_0h, k_0k]k_0kh_0k^{-1}k_0^{-1}h_0^{-1}\\
  &=[h_0, k_0][k_0, h_0h][h_0h, k_0k][k_0k, h_0]\in (H, K).
  \end{align*}
