Answer to Question #139354 in Abstract Algebra for J

Question

Assume that (G,*) is a group, that g is an element of G and that t is the smallest positive integer such that g^t=e

prove that g^n=e if and only if t |n. use division algorithm and explain why r must = 0

Answer 1

Let us prove the converse first.

If $t|n$ . thus there exist $k\in \Bbb Z$ such that $kt=n$ .

Now,

$$g^n=g^{kt}=(g^t)^k=e^k=e$$

Thus,

$$g^n=e$$

Suppose, $t$ not divides $n$ , hence by division algorithm there are p,r such that

$$n=pt+r,\hspace{1cm}0<r<t$$

Moreover we have given

$$g^n=e\\ \implies g^n=g^{pt+r}\\ \implies(g^t)^pg^r\\ \implies e^pg^r\\ \implies eg^r=g^r=e$$

But we have given that $t$ is given smallest , hence by definition $t|r\implies r\ge t$ , hence we get a contradiction.

Therefore r must 0, hence

$$n=pt\implies t|n$$

We are done.