# If we have $A \cap B =\{e \}$, then $ab=ba$?

Suppose we have sets $A$ and $B$. In group theory, is it always true that if we have $A \cap B =\lbrace e \rbrace$, then $ab=ba$?

This question relates to my study of the proof of 'if G is the internal direct product of $K_1,...,K_n$, then for any $i \neq j$, we have $k_i k_j = k_j k_i$'. The aim of the proof is to show that $K_i \cap K_j =\{ e \}$.

I am just curious if this is true for any arbitrary sets or just in internal direct products.

• Your claim is true if $A,B\triangleleft G$. That's likely the case in the proof you're addressing. I could explain it, but it's easier to see: crazyproject.wordpress.com/2010/03/14/… – Ian Coley Apr 16 '13 at 16:48
• What are $a,b$? – Thomas Andrews Apr 16 '13 at 16:53
• Please consider that this question doesn't make sense for arbitrary sets. You need at least an operation, possibly defined on some bigger structure containing both $A$ and $B$. – A.P. Apr 16 '13 at 16:54
• In a direct product, you have that $G=AB$, $A\cap B=1$ and $A, B\lhd G$. Then $ab=ba$ for all $a\in A$ and $b\in B$. There is a sliding scale of products where the first two conditions hold, $G=AB$ and $A\cap B=1$: Direct products, semidirect products ($A\lhd G$, $B<G$) and Zappa-Szep products ($A, B<G$). What you say holds in direct products, but will only hold in either of the other two if they are, in fact, direct products. So, for example, in amWhy's answer $S_3$ is a semidirect but not a direct product, so $ab=ba$ cannot hold for all $a\in A$ and $b\in B$. – user1729 Apr 16 '13 at 19:08
• (Although I should say that $G=S_3$ is the semidirect product of $A=\langle (1, 2)\rangle$ and $B=\langle (1,2,3)\rangle$. That is, $A<G$ $B\lhd G$, $AB=G$ and $A\cap B=1$. If we take $B=\langle (2,3)\rangle$ as in amWhy's answer then neither $A$ nor $B$ are normal and $AB\neq G$ (indeed, $AB$ is not even a subgroup of $G$!).) – user1729 Apr 16 '13 at 19:16

Consider the free group $F_2$ in two generators $a$ and $b$.

Then clearly $ab \ne ba$, but $\langle a\rangle \cap \langle b \rangle = \{e\}$. (Here $\langle a \rangle$ denotes the subgroup generated by $a$.)

• 16 and beyond my friend. +1 – mrs Jun 8 '13 at 10:53

No, it's not true for arbitrary groups/subgroups:

Assuming you're talking about subgroups/groups, even if two groups $H, K$ are subgroups of the same group $G$, we can have $H\cap K = \{e\}$, but $hk \neq kh$ for some $h \in H, k\in K$.

Consider the symmetric group $S_n$ and its subgroups. Fore example, if we have $S_3$ with $H \leq G, K \leq G$ with $H = \langle (1, 2)\rangle$ and $K = \langle (2, 3)\rangle$, with $G = \langle(1 2), (23)\rangle$.

Then $H \cap K = \{e\}$, but $(12)(23) \neq (23)(12)$

• Yeah, you answered my question – Idonknow Apr 16 '13 at 17:11
• Nice Amy. I love it ;-) – mrs Apr 17 '13 at 6:19
• @amWhy: I second or friends Babak's comments! :-) +1 – Amzoti Apr 18 '13 at 0:27

Or consider the infinite dihedral group $D_{\infty}$:

$$D_{\infty}=\langle a,b\mid a^2=b^2=1\}=\langle s,t\mid t^2=1,tst^{-1}=s^{-1}\rangle$$

• Another nice example (or I should say counterexample)! – Namaste Apr 18 '13 at 0:43