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Exhibit two subgroups which do not commute in Symmetric group S4

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.13

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 $S_4$.

Solution: 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 $HK \neq KH$.


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  1. I think this is incorrect. HK = KH = S_4, just by the order theorems. Also, HK = KH is not conditional on the elements of K commuting with the elements of H.

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