Abstract Algebra - Subgroup of Permutations

iamalexalright
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Homework Statement


A is a subset of R and G is a set of permutations of A. Show that G is a subgroup of S_A (the group of all permutations of A). Write the table of G.


Onto the actual problem:
A is the set of all nonzero real numbers.
G={e,f,g,h}
where e is the identity element, f(x) = 1/x, g(x) = -x, h(x) = -1/x

Would this be the right way to do it?

For each combination of elements in G (call the elements a,b) I need to show
a*b is in G

I also need to show that the inverse of a is in G.

Here is where I get confused, I'll start with with a = e:

ee = e
ef = f
eg = g
eh = h

Okay, that is all good, now letting a = f:
fe = e
ff = e
fg = h
fh = g

now a = g:
ge = g !
gf = h
gg = g !
gh = e

now a = h:
he = h
hf = g !
hg = f
hh = g !

What am I doing wrong here?
 
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The following are wrong:

fe=e
gg=g
gh=e
hg=f
hh=g

What did you do to obtain those?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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