Subsets of a group

We return again to the general case where $G$ is an arbitrary group. Let $S$ be an arbitrary nonempty subset of $G$, $S\subset G$ and $S\not= \emptyset$; such a set $S$ is usually called a complex of $G$. If $S_1$ and $S_2$ are two complexes of $G$, the product $S_1S_2$ is defined as follows:

$$S_1S_2=\{g_1g_2\in G\ \vert\ g_1\in S_1 \ {\rm and}\ g_2\in S_2\}.$$

If $S_1 = \{g\}$, a singleton set, we shall write $gS_2$ instead of $\{g\}S_2$. (A similar notation will be followed if $S_2$ is a singleton set.) It is clear, by the associative law, that if $S_1$, $S_2$, and $S_3$ are complexes of $G$, then

$$S_1(S_2S_3)=(S_1S_2)S_3.$$

Finally if $S$ is a complex of $G$, we denote by $S^{- 1}$ the following set

$$S^{-1}=\{ g^{-1}\ \vert \ g\in S\}.$$

Using the notation just introduced, we can characterize, according to exercise 2 of Chapter 2, a subgroup as follows: a nonempty subset $H$ of a group $G$ is a subgroup if and only if

$$H\cdot H^{-1}=H.$$ (4.4)

It is also clear that if $S = H \leq G$, then $H^2 = HH = H$, and $H^{-1} = H$. Next suppose that $H_1\leq G$ and $H_2 \leq G$. Assume that $H_1H_2 \leq G$. If $a_1 \leq H_1$ and $a_2\leq H_2$, then $a_1^{-1}\in H_1$ and $a_2^{-1}\in H_2$, therefore $a_1^{-1}a_2^{-1}\in H_1H_2$. But since $H_1H_2$ is assumed itself to be a subgroup, it must contain which is the general element of $H_2H_1$, i.e., we have shown that $H_2H_1 \subset H_1H_2$. Similarly,to show that $H_1H_2 \subset H_2H_1$, we need to show that a general element $a_1a_2$ of $H_1H_2$ is in fact in $H_2H_1$. Since $H_2H_1 \subset H_1H_2$, we have $(a_1a_2)^{-1}=a_2^{-1}a_1^{-1}=b_1b_2$, where $b_1\in H_1$, $b_2\in H_2$. This implies $a_1a_2=b_2^{-1}b_1^{-1}$, so $H_1H_2 \subset H_2H_1$. Together, these imply $H_1H_2 = H_2H_1$. Conversely, suppose that $H_1H_2 = H_2H_1$. Then $H_1H_2(H_1H_2)^{-1} =$ $H_1H_2H_2^{-1}H_1^{-1} =H_1H_2H_1=H_2H_1H_1=H_2H_1=H_1H_2$, hence $H_1H_2 \leq G$ from the characterization of subgroups given in (4.4). We, therefore, have established the following result.

Theorem 4.2.1   The product $H_1H_2$ of two subgroups $H_1$, $H_2$ of a group $G$ is itself a subgroup if and only if $H_1$ and $H_2$ commute, i.e., if and only if $H_1H_2 = H_2H_1$.

Warning: We caution the reader that when we say $H_1$ and $H_2$ commute, we do not demand that this is so elementwise. In other words, it is not demanded that $h_1h_2 = h_2h_1$ for all $h_1 \in H_1$ and all $h_2 \in H_2$; all that we demand is that for any $h_1 \in H_1$ and $h_2 \in H2$ $h_1h_2 =h_2'h_1'$, for some elements $h_1'\in H_1$ and $h_2'\in H_2$. For example, in $S_3$, if $H_1 = \{(1), (123), (132)\}$ and $H_2 = \{(1), (12)\}$ then as the reader can verify $H_1H_2 = S_3$ and $H_2H_1 = S_3$, so that $H_1H_2 = H_2H_1$. But note that $(1,2,3) (1,2) \not=(1,2) (1,2,3)$ whereas $(1,2,3) (1,2) = (1,2) (1,3,2)$.