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This question is Exercise 1.3 from Rotman's 1994 text, "An introduction to the theory of groups".

For all $\alpha, \beta, \gamma \in S_x$, prove that $\alpha(\beta\gamma) = (\alpha\beta)\gamma$. Indeed, if $X, Y, Z, W$ are sets and $f:X\rightarrow Y$, $g:Y\rightarrow Z$, and $h:Z\rightarrow W$ are functions, then $h(gf) = (hg)f$.

The four properties for groups are closure, associativity, identity, and inverse. The two exercises before 1.3 ask for identity and inverse for 1.1 and 1.2 respectively. They weren't all that difficult.

Using the definition of a permutation on p.2 as a bijection $\alpha:X\rightarrow X$, identity is shown by multiplying each element by 1, or 1 by it, which gives the original element.

Inverse was "proved" by simply assuming the property I required (by assuming another element, call it $\beta$ already existed and was the inverse of the permutation $\alpha$). This is was suggested by the author. I'm not happy with it because it really tested the property of inverses (that they cancel) and not that groups all have them. If someone could provide a proof that doesn't assume the conclusion that'd be great. But that's just a minor annoyance for now.

For the time being, I'd like to prove the associativity property. I can see that any symmetric group has a corresponding Cayley graph, and in calculating the elements any combination of them obeys the associativty property, but that's not a proof is it?

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$[h(gf)](x)=h([gf](x))=h(g(f(x)))=[hg](f(x))=[(hg)f](x)$. Every bijection on a set has an inverse: it’s an immediate consequence of being a bijection. –  Brian M. Scott Nov 1 '11 at 21:54
@bwkaplan: You are overthinking this. There's no need to invoke a Cayley graph. Just use the definition of the symmetric group as the set of bijections from a set to itself, as suggested by Brian M. Scott above and Alex Youcis below. –  Grumpy Parsnip Nov 1 '11 at 23:32

1 Answer 1

Not to be too blunt, but you do realize this is just the associativity of function composition? Just do it. For any $X\xrightarrow{f}Y\xrightarrow{g}Z\xrightarrow{h}W$ and any $x\in X$ one has that $(h\circ (f\circ g))(x)=h((g\circ f)(x))=h(g(f(x))$ and $((h\circ g)\circ f)(x)=(h\circ g)(f(x))=h(g(f(x))$ etc.

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yes, you caught me second-guessing myself. –  user4536 Nov 2 '11 at 14:17

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