If $H$ and $K$ are subgroups of $G$, show that $(H, K)$ is normal in $H \vee K$ Here the group $(H, K)$ is the group generated by elements of the form $hkh^{-1}k^{-1}$ with $h \in H$ and $k \in K$. 
This is a question in Hungerford's Algebra. 
It's pretty clear to me that we can show the desired result by showing that $H$ and $K$ are in the normalizer of $(H, K)$. However, it's not clear to me how to show this, that is, if $ c\in (H,K)$ why should $hch^{-1} \in (H, K)$. The trick used when showing that the commutator subgroup is normal fails as $hkh^{-1}$ need not be in $K$. It seems that other tricks used to show that the commutator subgroup is normal will fail as well considering that if $H$ and $K$ are not both abelian then $(H, K)$ is not the commutator of their join, and so isn't a typical commutator subgroup. 
Any hints would be welcome!
 A: We want to show that if $M$ and $N$ are subgroups of $G$ then $[M,N]$ is a normal subgroup of the group $\langle M, N \rangle$. What we will do is show that $\langle M, N \rangle$ is contained in the normaliser, $N_{G}([M,N])$. Since the normaliser is a subgroup (and so is closed), it is sufficient to show that $M$ and $N$ are in $N_{G}([M,N])$. 
First we will show that $M$ is in $N_{G}([M,N])$. This means that $M$ normalises $[M,N]$, in other words, we want to show that if we take $x \in [M,N]$ and $m \in M$ then $x^{m} \in [M,N]$. 
Any element $x$ of $[M,N]$ consists of a product of commutators and their inverses. Explicitly the element $x$ looks something like $[m_{1},n_{1}]^{\pm 1}\dots [m_{t},n_{t}]^{\pm 1}$ where $m_{i} \in M, n_{i} \in N$. We want to show that $x^{m} \in [M,N]$. Recall that a conjugate of a product is equal to a product of conjugates: $(ab)^{g}= a^{g}b^{g}$. Thus if we can show that for any $m\in M$ we have $[m_{i},n_{i}]^{m} \in [M,N]$ and $([m_{i},n_{i}]^{-1})^{m} \in [M,N]$ then it will follow that $x^{m} \in [M,N]$ and hence $M$ is in $N_{G}([M,N])$. 
Recall the commutator identity: $[a,b]^{c}=[ac,b][b,c]$.
Using this we see that $$[m_{i},n_{i}]^{m}=\underbrace{[m_{i}m,n]}_{\in[M,N]}\underbrace{[n_{i},m]}_{=[m,n_{i}]^{-1}}\in [M,N]$$
In a similar way we see that $([m_{i},n_{i}]^{-1})^{m}=([m_{i},n_{i}]^{m})^{-1}\in [M,N]$.
Hence it follows that for any $m \in M$ and any $x \in [M,N]$ we have $x^{m} \in [M,N]$ and hence $M \leq N_{G}([M,N])$. 
A similar argument works for $N$, using the commutator identity $[a,b]^{c}=[c,a][a,bc]$.
Thus both $M$ and $N$ are contained in $N_{G}([M,N])$ and hence you can conclude $\langle M, N \rangle \leq N_{G}([M,N])$.
