If you find any mistakes, please make a comment! Thank you.

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 $HK \neq KH$.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has One Comment

  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.

Leave a Reply

Close Menu