# Permutations (and isomorphisms)

#### quasar987

Homework Helper
Gold Member
I decided not to use this convention, because it gets confusing when you have permutations composed with non-permutations functions.. you then have to stare at the thing forever just to see what's in what.

So how would the proof go w/o this silly notation? It seems we still have to show the same thing, namely, that

$$h(a\cdot b) = h(a) \circ h(b)$$

since the definition of homomorphism is independant of the notation we use. But now the problem is genuine, since

$$h(a) \circ h(b) = h_a(gb)$$

#### matt grime

Homework Helper
you have to use the opposite notation or similar.

one way is to define h_a(g)=ag and then
h_ab(g) = abg

and h_a(h_b(g))=h_a(bg)=abg

another is to use the inverse operator and define h_a(g)=ga^{-1}

you cannot mix conventions and expect them to work like you did. the existence of the homomorphism doesn't depend on the notation, agreed, but what you wrote down is not a homomorphism with the mixed conventions you used.

Last edited:

#### quasar987

Homework Helper
Gold Member
Oh right, adjust the definition of h_a, great!

#### dimuk

"I from S_{n-1} to S_n that sends f in S_{n-1} to the g in S_n g(r)=f(r) for 1<=r<=n-1 and g(n)=n is clearly a bijection between S_{n-1} and the permutations that fix n in S_n since it is invertible, the inverse being the function muzza gave in the last post but one. "

Can u explain this more.

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