1. Automorphism Proof

Attempt:

I don't know how to show that \pi (g) = g^{-1}, using the given hint. Any help to get me started is really appreciated. But I have tried to show that G is abelian and that it has an odd order:

If $\displaystyle \pi$ is an automorphism then for all $\displaystyle g,h \in G$

Hence G is Abelian. (Right?)

Now to show that the group is of an odd order. We can split the whole set G\{1} into disjoint pairs .

Then G\{1} has even number of elements and thus $\displaystyle G$ (including 1) has an odd number of elements.

Is this right?

P.S. Latex wasn't working properly today.

2. Originally Posted by demode

Attempt:

I don't know how to show that \pi (g) = g^{-1}, using the given hint. Any help to get me started is really appreciated. But I have tried to show that G is abelian and that it has an odd order:

If $\displaystyle \pi$ is an automorphism then for all $\displaystyle g,h \in G$

Hence G is Abelian. (Right?)

Now to show that the group is of an odd order. We can split the whole set G\{1} into disjoint pairs .

Then G\{1} has even number of elements and thus $\displaystyle G$ (including 1) has an odd number of elements.

Is this right?

P.S. Latex wasn't working properly today.
I would actually do it in reverse thought (I've done this problem before, Dummit and Foote--right?). Namely, the key thing to note is that to show "a=b" here you just need to show that "pi(ab')=ab' " where ' denotes inverse. For example, fiddle around to show that pi(aba'b')=aba'b' and conclude that aba'b'=e. Then, using abelianess note that pi(pi(g)g)=pi(pi(g))pi(g)=gpi(g)=pi(g)g and so conclude that pi(g)g=e for every g in G. The rest of your work looks fine.

3. if π(a)a^-1 = π(b)b^-1, then π(b^-1a) = b^-1a, whence b^-1a = e, so a = b.

this shows we can write any element of G uniquely as π(a)a^-1 for some a.

so for any g in G, gπ(g) = (π(a)a^-1)(π(π(a)a^-1)) = (π(a)a^-1)(aπ(a^-1)) = π(a)π(a^-1) = π(e) = e.

hence π(g) = g^-1, for all g in G.

but if g-->g^-1 is an automorphism: then (gh)^-1 = h^-1g^-1 = g^-1h^-1, so hg = gh.

if G has an element x of order 2, then π(x) = x^-1 = x, and x is a fixed point of π, so G has no elements of order 2,

and hence G must be of odd order.