Fix any labeling of the vertices of a square and use this to identify as a subgroup of . Prove that the subgroups and do not commute in .

We can label the vertices of a square as follows.

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

Now let and . If , then in particular , a contradiction. So .

## Comments

A typo in the last sentence. $HK \neq KH$.

And I have a question. I thought the problem was asking that the elements of and commute, i.e. for all and , which is different from that

Thanks.

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 and commute, which I interpret to mean that the subgroups commute, rather than that the elements commute pairwise. That is, and commute if for all and , there is

someelement of such that .I do think there are good reasons to interpret this in both ways; ‘pairwise’ commutativity is relevant when showing that two subgroups generate an internal direct product, and ‘setwise’ commutativity is relevant when showing that one subgroup normalizes another.