The Product of Subgroups of an Abelian Group

Reference: Fraleigh p. 58 Question 5.43 in A First Course in Abstract Algebra

Let $G$ be an Abelian Group. Suppose $H$ and $K$ are subgroups of $G$, and $HK = \{ xy: x \in H \text{ and } y \in K \}$. Prove that $HK$ is a subgroup of $G$.

Since $H$ and $K$ are both subgroups, $e \in H$ and $e \in K$. $ee \in HK$ which shows that $HK$ is not empty.

(i). Let $ab, cd \in HK$ such that $a, c \in H$ and $b, d \in K$. First I need to show that $abcd$ can be written as a product of elements of $H$ and elements of $K$. Since $G$ is Abelian, $abcd = acbd = (ac)(bd)$. We know that $(ac) \in H$ and $(bd) \in K$.

(ii). Let $ab \in HK$, then $a \in H, b \in K$. I need to show that $(ab)^{-1} \in HK$ can be written as a product of elements of $H$ and $K$. Because $a \in H$ and $H$ is a subgroup, $a^{-1} \in H$. And since $b \in K$ and $K$ is a subgroup, $b^{-1} \in K$. $(ab)^{-1} = b^{-1}a^{-1}$.

Am I interpreting the question correctly? And is my proof correct?

@Arturo's Exercise:

I'm a bit confused as to the definition of $HK = KH$ given. I thought that if two groups $A$ and $B$ were equal, every element of $A$ would be contained in $B$ and every element of $B$ would be contained in $A$.

(i). Let $ab, cd \in HK$, then $abcd \in HK$. I need to show that $abcd \in HK$. Using the definition above, $ab = b'a'$ and $ab = a''b''$ where $a', a'' \in H$ and $b', b'' \in K$. A similar statement can be made for $cd$.

(ii). Let $ab \in HK$, then $(ab)^{-1} \in HK$

Could I get a hint as to how to start part (i)? I've tried substituting to try and show that $abcd$ is a product of elements of $H$ and $K$, but I have not gotten anywhere.

• Yes, your proof is fine (strictly, you could make it more complete by explicitly noting that $b^{-1}a^{-1} = a^{-1}b^{-1}$ as $G$ is abelian, putting the element in the right form to be in $HK$). Commented Aug 12, 2011 at 20:51
• @Jon: It seems you misunderstood/are confused about the "exercise" I wrote below. If $A$ and $B$ are subsets of $G$, then $AB$ is the subset of all products of the form $ab$ with $a\in A$, $b\in B$. For example, in $S_3$, if $A=\{(1,3), (1,2,3)\}$ and $B=\{(1,2),(1,3)\}$, then $AB=\{(1,3)(1,2), (1,3)(1,3), (1,2,3)(1,2), (1,2,3)(1,3)\} = \{(1,2,3), I, (1,3), (2,3)\}$. If $H$ and $K$ are subgroups, then they are also subsets, so you can consider the set $HK$. It may or may not be a sub group. The exercise is to show exactly when it is a subgroup, and it is a set-theoretic condition. Commented Aug 16, 2011 at 16:45

The last part, as Geoff notes, is perhaps a bit lacking: you did not show that $$b^{-1}a^{-1}$$ is an element of $$HK$$, because that requires you to show that it can be written as the product of something in $$H$$ times something in $$K$$, rather than "something in $$K$$ times something in $$H$$" (yes, they are the same because $$G$$ is abelian, but this needs to be said somewhere).

Now, for further practice, you can try the usual problem:

Let $$G$$ be a group, not necessarily abelian, and let $$H$$ and $$K$$ be subgroups. Prove that $$HK=\{ hk\mid h\in H,\ k\in K\}$$ is a subgroup of $$G$$ if and only if $$HK$$ and $$KH$$ are equal as sets.

(Note that $$HK=KH$$ means that for each $$h\in H$$ and $$k\in K$$ there exist $$h',h''\in H$$ and $$k',k''$$ in $$K$$ such that $$hk=k'h'$$ and $$kh = h''k''$$. We do not require $$hk=kh$$ for each $$h\in H$$ and $$k\in K$$.)

Let $$G$$ be a group. If $$A$$ and $$B$$ are subsets of $$G$$, then we can form the subset $$AB$$, $$AB = \{ab\mid a\in A, b\in B\}.$$

For example, take $$G=S_3$$, $$A=\{(1,2), (1,3)\}$$, $$B=\{(1,2,3),(1,3,2)\}$$. Then $$AB = \{(1,2)(1,2,3), (1,2)(1,3,2), (1,3)(1,2,3), (1,3)(1,3,2)\} = \{(2,3), (1,3), (1,2)\}.$$

If $$H$$ and $$K$$ are subgroups of $$G$$, then $$HK$$ may or may not be a subgroup (it's certainly a subset). For instance, if $$G=S_3$$, $$H=\{I,(1,2)\}$$, $$K=\{I,(1,3)\}$$, then \begin{align*} HK &= \{II, I(1,3), (1,2)I, (1,2)(1,3)\}\\ &= \{I, (1,3), (1,2), (1,3,2)\}, \end{align*} which is not a subgroup, since it is not closed under products or inverses. On the other hand, if $$H=\{I,(1,2)\}$$ and $$K=\{I, (1,2,3), (1,3,2)\}$$, then \begin{align*} HK &= \{II, I(1,2,3), I(1,3,2), (1,2)I, (1,2)(1,2,3), (1,2)(1,3,2)\}\\ &= \{I, (1,2,3), (1,32), (1,2), (2,3), (1,3)\}\end{align*} which is a subgroup (in fact, it's the entire group).

Now, in the first example above, with $$H=\{I, (1,2)\}$$ and $$K=\{I, (1,3)\}$$, we can also construct $$KH$$. We have: \begin{align*} KH &= \{II, I(1,2), (1,3)I, (1,3)(1,2)\}\\ &= \{I, (1,2), (1,3), (1,2,3)\}.\end{align*} Notice that $$KH$$ is not equal to $$HK$$: $$KH$$ contains $$(1,2,3)$$, which is not in $$HK$$.

Now look at the second example above, with $$K=\{I, (1,2,3), (1,3,2)\}$$ and $$H=\{I, (1,2)\}$$. We have: \begin{align*} KH &= \{ II, I(1,2), (1,2,3)I, (1,2,3)(1,2), (1,3,2)I, (1,3,2)(1,2)\}\\ &= \{I, (1,2), (1,2,3), (1,3), (1,3,2), (2,3)\}.\end{align*} Here, $$KH$$ is equal to $$HK$$. However, also notice that it is not true that $$hk=kh$$ for each $$h\in H$$ and $$k\in K$$. $$(1,3)$$ is in both $$HK$$ and in $$KH$$, but $$(1,3)$$ appears in $$HK$$ as the product $$(1,2)(1,3,2)$$, whereas it appears in $$KH$$ as the product $$(1,2,3)(1,2)$$ (different $$h$$).

The proposition I suggest is asking you to prove that if the set $$HK$$ is equal to the set $$KH$$, then $$HK$$ is a subgroup; and conversely, that if $$HK$$ is a subgroup, then the set $$HK$$ must be equal to the set $$KH$$.

Now, what are you assuming in (i)? That the sets are equal, or that $$HK$$ is a subgroup?

If we assume that $$HK$$ is a subgroup, that means that (i) it contains $$e$$; (ii) it is closed under products; and (iii) it is closed under inverses. Your objective is to show that $$HK=KH$$. In order to show that $$HK=KH$$, you need to show that $$HK\subseteq KH$$ and that $$KH\subseteq HK$$.

So, in order to show that $$HK\subseteq KH$$, you need to show that if you have $$x\in HK$$, then $$x\in KH$$. So, let $$x\in HK$$. That means that there exist $$h\in H$$ and $$k\in K$$ such that $$x=hk$$. What you need to show is that $$x\in KH$$; that is, you need to show that there exists $$k'\in K$$ and $$h'\in H$$ (it's possible that $$k=k'$$, but then again it's possible that $$k\neq k'$$; we don't know!) such that $$x=k'h'$$.

Well... we know that $$HK$$ is a subgroup of $$G$$. Since $$x\in HK$$, then we must have $$x^{-1}\in HK$$; that means that there exist $$h''\in H$$ and $$k''\in K$$ such that $$x^{-1}=h''k''$$. Therefore...

Now finish this part of the argument, and then show that $$KH\subseteq HK$$.

Conversely, you need to show that if $$HK=KH$$, then $$HK$$ is a subgroup. You need to show that: (i) $$e\in HK$$ (easy, since $$H$$ and $$K$$ are both subgroups); (ii) that if $$x,y\in HK$$, then $$xy\in HK$$; and (iii) that if $$x\in HK$$, then $$x^{-1}\in HK$$.

Well, for (iii), for example: suppose that $$x\in HK$$. Since $$x\in HK=KH$$, then there exist $$h\in H$$ and $$k\in K$$ such that $$x=kh$$. What is $$x^{-1}$$? Is it in $$HK$$? Etc. Finish this off.

• Look above at my question ^ Commented Aug 16, 2011 at 16:45