- #1

penroseandpaper

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The textbook asks me to use subgroup axioms to prove why a set of permutations that interchange two specific symbols in S4 is or isn't a subgroup of the symmetric group, and the same for a set of permutations that fix two elements.

My guess is that the set of permutations that interchange the two symbols isn't a subgroup under such rules because it doesn't contain the identity element. Meanwhile, the one that fixes the two symbols does contain the identity element and hence satisfies axiom 2.

I was wondering if I'm right in saying that, and whether I need to consider either of the two other axioms in proving it. My thoughts are it's otherwise associative and inverses contained.

My calculations also produced four permutations in cycle form for the interchanging set and two permutations for fixing (identity and one more) - did I find them all?

Group theory as a lockdown challenge is proving a little trickier than expected! But it doesn't help that so many textbooks don't have any answers in them...

Still, it's nice to stretch the grey matter.

Sorry to bother you and thanks for your help,

Penn