# Thread: Number of elements in a group

1. ## Number of elements in a group

Let G be an abelian group of order 10. Let S={g in G | g^(-1) = g }
Then what is the number of non identity element sis S?

I think the anwer is 0.
Because We cn observe that S is a normal subgroup of G.

I define a map f : G -> G given by
f(g) = g^2 for all g in G
This is an onto homomorphism with kernel S.
So G/S ~ G
So o(G)/o(S) = o(G)
So S has just one element 'e'

Am i right.
just have one doubt. Is the homomorphism srjective??

2. Originally Posted by poorna
Let G be an abelian group of order 10. Let S={g in G | g^(-1) = g }
Then what is the number of non identity element sis S?

I think the anwer is 0.
Because We cn observe that S is a normal subgroup of G.

I define a map f : G -> G given by
f(g) = g^2 for all g in G
This is an onto homomorphism with kernel S.
So G/S ~ G
So o(G)/o(S) = o(G)
So S has just one element 'e'

Am i right.
just have one doubt. Is the homomorphism srjective??

There is at least one non-identity element in S: to see this, pair up every element in G with its inverse....what do you get?

By the way, the map $f(g)=g^2$ is a homomorphism on $G$ iff $G$ is abelian, so you can use it since you assume the group is abelian...but the result still is true if the group isn't abelian: there's always an element (one, at least) of order 2 in any finite group of even order.

Tonio

3. Can I not determine exactly the number of elements in S? I have been asked to.
I can see now that by pairng every elment with their inverse, S is non trivial. So o(S)|o(G)

So S should have 2 or 5 elements right?

4. Isn't an abelian group of square-free order cyclic?

5. Oh is that so? Then it becomes easy The order is 2. Thanks