# Thread: a group that is a subgroup of Zn

1. ## a group that is a subgroup of Zn

For each odd composite numbern=>3, any number satisfying a^(n-1)≡1(modn )
is said to be a Fermat liar for n.
F-Liar={ a/1<= a<= n and a^(n-1)≡1(modn }
Show that F-Liar is a subgroup of Z^*n
i tried to solve this by finding the size or order of F and it must divide the order of Zn to be F a subgroup of Zn.
I think the size of Zn is n-1
and the size of F is n
but i could not get the whole prove
the order of Zn is less than or equal n/2 and the order of F must be less than or equal the order of Zn but how can I show that.

2. Originally Posted by koko2009
For each odd composite numbern=>3, any number satisfying a^(n-1)≡1(modn )
is said to be a Fermat liar for n.
F-Liar={ a/1<= a<= n and a^(n-1)≡1(modn }
Show that F-Liar is a subgroup of Z^*n
i tried to solve this by finding the size or order of F and it must divide the order of Zn to be F a subgroup of Zn.
I think the size of Zn is n-1
and the size of F is n
but i could not get the whole prove
the order of Zn is less than or equal n/2 and the order of F must be less than or equal the order of Zn but how can I show that.

This is a mess: first, $\mathbb{Z}_n^*$ is the multiplicative group of all units in $\mathbb{Z}_n$ and its order is $\phi(n)$ = Euler's Totient Function.
Second, you need to show that the product of any two Fermat liars is again a F.L., which is pretty easy since everything's abelian here.

Tonio