Group theory question!! order and subgroups!! Please help!!

Let G be a group and let a in G such that |a| = n

(a) Show that |g a g^-1| = n for all g in G

(b) Is the subset G_n = {g in G : g^n = e} a subgroup of G? Justify your answer (Hint: Consider G = S_3)

(a) note that (gag^-1)^n = (gag^-1)(gag^-1)....(gag^-1) (n times)

= ga(g^-1g)a(g^-1g)a.....(g^-1g)ag^-1 = g(a^n)g^-1

= geg^-1 = gg^-1 = e.

thus |gag^-1| divides |a|.

now suppose that (gag^-1)^k = e, for 0 < k < n.

this means that g(a^k)g^-1 = e, so

(a^k)g^-1 = g^-1, and thus

a^k = e, contradicting that |a| = n.

(b) no. the example of S3 will do nicely:

the elements g of S3 such that g^2 = e are:

{e, (1 2), (1 3), (2 3)}

this cannot possibly be a subgroup of S3, for one, it is not closed under multiplication, and for two, 4 does not divide 6.

Would you give me a little bit more explanation on part (b) please?

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