product of subgroups is in general not a subgroup

Question

Be $G$ a group with subgroups $U$ and $V$ and be $UV := \left\{uv \mid u \in U, v \in V \right\}$

I want to prove the following:

$UV$ is in general no subgroup of $G$.

Proof. Be $U \nsubseteq V$ and $V \nsubseteq U$. Be $x \in U \setminus V$ and $y \in V \setminus U$.

Hence $xy \in UV$ but $xy \notin U \cup V$. This implies $\left(xy \right)\left(xy\right) \notin UV$ per definition of $UV$, hence $UV$ is no subgroup of $G \quad \square$

I've heard that 'This implies $\left(xy \right)\left(xy\right) \notin UV$ per definition of $UV$' is not valid. I don't understand why this should be the case.

Would be $\left(xy \right)\left(xy\right) \in UV$, $xy$ had to be an element of $U$ or of $V$ per definition of $UV$. But it is $xy \notin U \cup V$.

What am I missing?

Answer 1

Given two subgroups $U$ and $V$, the set $UV$ may in fact be a group given certain conditions.

Some necessary and sufficient conditions: $UV$ is a subgroup of $G$ if and only if $UV = VU$.

Some sufficent conditions: If $U$ is contained in the normalizer of $V$, then $UV$ is a subgroup of $G$.

It would be a good exercise to show these two results. The second follows as a corollary of the first.

So in summary, $UV$ can be a group, but sometimes it's just a set. That is why a proof like the one you tried to write will not work: it is not sufficient to assume neither is contained in the other to show $UV$ is not a group.

Answer 2

In order to prove that a statement is not generally true, you ultimately need to show the existence of a counterexample to the statement. While your statements that try to pick apart where exactly the statement fails to hold are useful for proving partial results, they don't necessarily help here.

Here's a much quicker and much more correct proof: there exist subgroups $U$ and $V$ of $S_3$ for which $UV$ is not a subgroup. Find such subgroups, and show that their product is not a subgroup.

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As for why your statement was wrong, you wrote that if $x \in U$ and $y \in V$, then $xy \in UV$ and "this implies, per definition of $UV$, that $(xy)(xy) \notin UV$".

Let's look at an example where your statement doesn't hold: consider the group $\Bbb Z_2 \times \Bbb Z_2$ with the group product of modular addition. Let $U = \{(0,0),(1,0)\}$ and $V = \{(0,0),(0,1)\}$. Note that $(1,0) + (0,1) = (1,1)$ and that indeed, we have $(1,1) \notin U$ and $(1,1) \notin V$. However, $$ (1,1) + (1,1) = (0,0) \in UV $$ In fact, we find in this case that $UV$ is a subgroup since $UV = \Bbb Z_2 \times \Bbb Z_2$.

Answer 3

One clear way that $UV$ may fail to be a subgroup is that not every $vu$ can be written as $u'v'$.

This is related to $U$ being normal.

So, let's take a non-normal subgroup and $v$ of order $2$ to makes things easier.

The simplest example that comes to my mind is $G=S_4$, $U=S_3$, $V=\langle (14) \rangle$.

Then $vu=(14)(123) \in VU$ but is not of the form $u'v'$ because $(14)(123)(14)^{-1}=(14)(123)(14)=(234) \notin S_3$.