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Let $G$ be a multiplicatively written group and let $g\in G$.

Show that $g^k=e$ if and only if the order of $g$ is finite and divides $k$.

Solution

Suppose $\operatorname{ord}(g)$ divides $k$ and let $\operatorname{ord}(g)=n.$

So $k=tn$ for some $t\in\mathbb{Z}.$

Thus $g^k=g^{tn}=(g^{n)^t}=e.$

Now suppose $g^k=e.$

Where do I go from here?

Edit:

Suppose $g^k=e$ and suppose $\operatorname{ord}(g)=n.$

Now $k=nq+r$ for some $0\leq r<n.$

This now gives $e=g^k=g^{nq+r}=g^{nq}g^r=g^r.$ However, we have a contradiction as $n$ is the smallest integer for which $g^r=e.$ Thus $r=0$ and so $n$ divides $k$.

Is this correct?

MHW
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3 Answers3

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"Let $\operatorname{ord}(g)=n$, $g^k=e$ and assume that $k$ is not a multiple of $n$" is how you could continue. You use that assumption to construct an $m$ with $0<m<n$ such that $g^m=e$, which contradicts the definition of $n$.

Arthur
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  • I've made an edit, does this look something like what you're saying? – MHW May 09 '16 at 13:28
  • The way you're phrasing it doesn't give a contradiction, per se (you just show that $r=0$, which means $k=nq$), but it's correct, and essentially what I had in mind. – Arthur May 09 '16 at 13:30
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Consider $A= \{ k \in \mathbb Z : g^k = e \}$.

Then $A$ is a subgroup of $\mathbb Z$ and so $A=m\mathbb Z$ for some $m$.

Therefore, $ord(g)$ is finite iff $A\ne 0$ and $m=ord(g)$.

lhf
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Suppose $g^k=e$. From the order's definition, $ord(g)=m$ is finite and $m \le k$. Let $H_1 = \{e, g, g^2, .. ,g^{m-1}\}$ and $H_2 = \{e, g, g^2, .. ,g^{k-1}\}$. Then $H_1 \subset H_2 $ and $H_1$ is a subgroup of $H_2$, therefore $ord(H_1)$ divides $ ord(H_2)$