## Exhibit two subgroups which do not commute in Sym(4)

Fix any labeling of the vertices of a square and use this to identify $D_8$ as a subgroup of $S_4$. Prove that the subgroups $D_8$ and $\langle (1\ 2\ 3) \rangle$ do not commute in $S_4$.

We can label the vertices of a square as follows.

Now a 90 degree clockwise rotation corresponds to the permutation $(1\ 2\ 3\ 4)$ and a reflection across the (1,3) axis to $(2\ 4)$.

Now let $H = \langle (1\ 2\ 3\ 4), (2\ 4) \rangle$ and $K = \langle (1\ 2\ 3) \rangle$. If $HK = KH$, then in particular $(1\ 2\ 3\ 4)(1\ 2\ 3)(1\ 4\ 3\ 2) = (2\ 3\ 4) \in K$, a contradiction. So $HK \neq KH$.

A typo in the last sentence. $HK \neq KH$.
And I have a question. I thought the problem was asking that the elements of $D_8$ and $\langle (1\ 2\ 3) \rangle$ commute, i.e. $xy=yx$ for all $x \in D_8$ and $y \in \langle (1\ 2\ 3) \rangle$, which is different from that $HK=KH$
About the question, I remember being unsure about how to interpret this exercise (D&F, 3.2 #13). The original asks whether the ‘elements’ of $D_8$ and $\langle (1\ 2\ 3) \rangle$ commute, which I interpret to mean that the subgroups commute, rather than that the elements commute pairwise. That is, $A$ and $B$ commute if for all $a \in A$ and $b \in B$, there is some element $a^\prime$ of $A$ such that $ab = ba^\prime$.