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Let $A$ and $B$ be two subgroups of the same group $G$. let

$$AB=\{ab\,|\, a\in A,\, b\in B \}$$ and $$\langle A,B\rangle$$ the subgroup generated by $A$ and $B$.

Are $AB$ and $\langle A,B\rangle$ the same as sets? My guess is no since the element $aba\in \langle A,B\rangle$ but it is not in $AB$, but I guess when $A$ and $B$ commute: $AB=BA$ then this is true. is this correct?

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Yes, your intuition is quite right. – Grumpy Parsnip Jul 29 '11 at 22:47
So now you need to produce an explicit example where the two sets are not equal. As you have observed, $G$ cannot be Abelian. There is good reason to try to work with small $A$ and $B$, but not trivially small. – André Nicolas Jul 29 '11 at 23:03
up vote 13 down vote accepted

Yes, the set $AB$ is a subgroup of $G$ if and only if $AB = BA$, as can be found in many algebra texts, such as Herstein's "Topics in Algebra". It is certainly necessary that $AB = BA$, since $(ab)^{-1} = b^{-1}a^{-1}$ for $a \in A, b \in B$ and subgroups are closed under taking inverses. On the other hand, if we do have $AB = BA$, then for any $a,c \in A$ and $b,d \in B$, we have $(ab)(cd) = a(bc)d = a(c^{*}b^{*})d$ for some $c^{*} \in A$, $b^{*} \in B$, so that $(ab)(cd) = (ac^{*})(b^{*}d) \in AB$. Hence $AB$ is closed under taking inverses, and closed under the group operation, so $AB$ is a subgroup.

Also $AB = BA$ implies that $\langle A,B \rangle \subseteq AB \subseteq \langle A,B \rangle$, so $\langle A,B \rangle = AB$. Conversely, if $\langle A,B \rangle = AB$ (as a set), then $AB$ is a subgroup, so $AB = BA$.

By the way, if $AB = BA$, it is usual to say that the subgroups $A$ and $B$ are permutable, rather than that they commute. Note that permutability need not imply that all elements of $A$ commute with all elements of $B$.

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Note that, regardless of whether $AB$ is a subgroup or not, if both $A$ and $B$ are finite, then the number of elements it has is $$|AB| = \frac{|A||B|}{|A\cap B|}.$$

To see this, consider the map $A\times B \to AB$ given by $(a,b)\mapsto ab$. The map is clearly onto. If $x\in A\cap B$, then for each $a\in A$ and $b\in B$ you have $ax\in A$, $x^{-1}b\in B$, so $(ax,x^{-1}b)\in A\times B$ has the same image as $(a,b)$. Thus, each element of $AB$ is the image of at least $|A\cap B|$-many elements. Conversely, if $(a,b)$ and $(a',b')$ have the same image under this map, then $ab=a'b'$, hence $(a')^{-1}a=b'b^{-1}\in A\cap B$, and $a = a'(a'^{-1}a)$ and $b=(a'^{-1}a)^{-1}b' = (bb'^{-1})b'$, so any two pairs that map to the same element arise from an element of $A\cap B$. Thus, each image occurs $|A\cap B|$ times, so $|AB||A\cap B| = |A\times B|=|A||B|$.

In particular, take $G=S_3$, $A=\langle(1,2)\rangle$, and $B=\langle (1,3)\rangle$. Then $A\cap B = \{e\}$, so the number of elements in $AB$ is $|A||B|=4$. This cannot be a subgroup of $S_3$, by Lagrange's Theorem. So $AB$ is properly contained in $\langle A,B\rangle$ (as sets).

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Notice that this argument extends shows that if the finite group $G$ is not nilpotent, then it has a pair of Sylow subgroups which are not permutable. For there is a prime $p$ such that the Sylow $p$-subgroup of $G$ is not normal. Hence $G$ has two distinct Sylow $p$-subgroups $A$ and $B$, say. Then $AB$ is not a subgroup of $G$, since, the formula pointed out by Arturo shows that its cardinality is a power of $p$ greater than $|A|$, whereas $A$ is a $p$-subgroup of $G$ of maximal order. – Geoff Robinson Jul 30 '11 at 0:23
@GeoffRobinson: Sorry for this old question. I have a question. According to what you answered, in group $\mathbb Z\times\mathbb Z$, the subgroup generateb by $(1,2)$ and $(3,2)$ should be $<(1,2),(3,2)>=<(4,4)>$? If so is this true that $(\mathbb Z\times\mathbb Z)/<(4,4)>\cong \mathbb Z_4\times\mathbb Z_4$? Thanks. – Babak S. Oct 22 '12 at 19:57
My comment was only about finite groups. – Geoff Robinson Oct 22 '12 at 20:02
@Arturo Magidin , this proof is so nice :) the proof which i know uses lagrange theorem and count the numbers of left cosets of B ,namely , $aB$ where $a \in A$. i liked your proof more ! greetings ! – Maths Lover Jun 13 '13 at 12:44

Let $G$ be $S_3$, the group of all permutations on $3$ letters.

Let $A$ be the two-element subgroup generated by the transposition $(1,2)$, and $B$ the two-element subgroup generated by $(1,3)$. Then $AB$ consists of the identity, $(1,2)$, $(1,3)$, and their product $(12)(13)=(1,2,3)$.

But $\langle A, B\rangle$ is all of $S_3$. We can verify this directly. Or else note that $\langle A, B\rangle$ has at least $4$ elements. Since the order of $\langle A, B\rangle$ must divide $3!$, $\langle A, B\rangle$ is of $S_3$.

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