# Showing Equality in terms of commutators

Let $$G$$ be a group and $$H=\langle a_1,a_2,\ldots, a_r\rangle$$ and $$K=\langle b_1,b_2,\ldots,b_s\rangle$$ be two its normal subgroups. Suppose $$M=\langle a_1,a_2,\ldots, a_r, b_1,b_2,\ldots,b_s\rangle.$$

My question is that can we prove that $$M=AB$$, where $$A=\langle a_i,[a_j,b_k], 1\leq i,j\leq r, 1\leq k\leq s\rangle$$ and $$B=\langle b_i,[b_j,a_k], 1\leq i,j\leq s, 1\leq k\leq r\rangle$$, where $$[a,b]=aba^{-1}b^{-1}$$?

• I am considering both $A$ and $B$ as normal subgroups, but can't see the equality, please help me to show this.
– MANI
Jan 11, 2022 at 14:35
• I don't see the problem. Since $H$ and $K$ are both normal, we have $M=HK$, and $H \le A \le M$, $K \le B \le M$, so $M=AB$. Jan 11, 2022 at 14:44

Lemma. Let $$G$$ be a group and $$H,K\subseteq G$$ two subgroups such that one of them is normal. Then $$HK=\langle H\cup K\rangle$$.

Proof. So "$$\subseteq$$" clearly holds. Always.

WLOG assume that $$K$$ is normal, i.e. $$gK=Kg$$ for any $$g\in G$$. Let $$x\in\langle H\cup K\rangle$$. Then there are $$a_1,\ldots, a_n\in H$$ and $$b_1,\ldots,b_n\in K$$ such that

$$x=a_1b_1\cdots a_nb_n$$

By the $$gK=Kg$$ equality we can replace every $$ab$$ by $$\tilde{b}a$$ (and vice versa), and then by induction we can rewrite $$x$$ as:

$$x=a_1'\cdots a_n'b_1'\cdots b_n'$$

for some $$a_1',\ldots,a_n'\in H$$ and $$b_1',\ldots,b_n'\in K$$, meaning $$x\in HK$$. $$\Box$$

Now note that both $$A=H$$ and $$B=K$$, simply because $$A$$ is generated by all $$a_i$$, and adding $$[a,b]=aba^{-1}b^{-1}$$ doesn't do anything, since $$ba^{-1}b^{-1}\in H$$ because it is normal, and therefore $$[a,b]\in H$$. The same applies to $$B$$ and $$K$$.

Thus by our lemma we get that $$AB=\langle A\cup B\rangle$$, where the right side is clearly $$M$$.