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.

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Comments

  • Gobi Ree  On November 24, 2011 at 9:26 pm

    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

    • nbloomf  On November 24, 2011 at 10:09 pm

      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 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.

      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.

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