
False witnesses
I'm stuck on one part of a problem.
"suppose $\displaystyle \Phi(n)=2t$ where t is odd, and $\displaystyle n=p^e$ for odd prime p and positive integer e.
Let L be the subgroup of false witnesses, i.e. L is the set of congruence classes made up of integers b prime to n such that $\displaystyle b^{(n1)/2)}\equiv \left(\frac{b}{n}\right)\ mod n$
Now suppose the order of L is $\displaystyle L=\Phi(n)/2=t$.
Let g be a primitive root mod n,so g has order $\displaystyle \Phi(n)$ and every element prime to n is of the form $\displaystyle g^j mod n$ for some j.
Show that for every integer $\displaystyle g^j$ belonging to a congruence class in L, j must be even. Since L=t, explain why $\displaystyle L=g^{2i}: i=1....t$"
I really don't know what to do here so any help would be appreciated.

Are you sure you posted the problem exactly?

Sorry, the question is a little confusing, but everything I typed is correct, but I did leave a part out at the beginning,
"$\displaystyle n=p^e$, suppose $\displaystyle p1=2s$ where s is an odd integer. Show that $\displaystyle \phi(n)=2t$, where t is odd."
I think I did that part correctly. The bulk of the problem is trying to show that
Suppose it happens that $\displaystyle L=\phi(n)/2=t$.
is impossible through 3 steps, so it's kind of a guided proof by contradiction (or so i think).
1. The first step is to explain why every element of L must have odd order. I think I did this part correctly but I'm not 100% positive.
2. The next part is the question about the primitive g mod n.
3. Show that $\displaystyle g^{n1} \equiv 1 mod n $, and use that to show $\displaystyle \phi(n)(n1)$. Why is this impossible.
So I thought the actual contradiction comes from part three, which I haven't even gotten to yet.