# If $g$ is an element of order $d$ and $d$ divides $n$, then $gn = 1$

Prove if $g$ is an element of order $d$ and $d$ divides $n$, then $gn = 1$.

I need help on how to prove the converse which comes from this theorem that I solved.

Suppose $gn = 1$ in a group $G$ and let $d$ be the order of $g$. Then $d$ divides $n$.

Proof: By the division algorithm, we may write $n = dq + r$ with $0 < r < d$. Then

$$g^r = g^{n-dq}=g^n=(g^d)^{-q}=1 \times 1=1.$$

If r > 0, this contradicts minimality of the order $d$. Hence $r = 0$ and so $d$ divides $n$.

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Why did you suppress the things you tried to solve this? This makes for an extremely bad question. –  Did Apr 3 at 17:36

Hint:

Divide $\,n\,$ by $\,d\,$ with residue:

$$n=xd+r\;,\;\;r=0\,\,\vee |r|<d\implies 1=g^n=g^{dx+r}=(g^d)^xg^r=1\cdot g^r=g^r\implies\ldots$$

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He already proved that. He's asking about the easier converse, see his comment to my answer. –  Math Gems Apr 3 at 16:56
The last part of his post says "Hence r= 0 and so d divides n"...I think the OP confused the question as he just posted a few minutes ago a question asking about the direction you show here (and I also answered that other question)... –  DonAntonio Apr 3 at 16:59
+1 for one of my teacher here. –  B.S. Apr 3 at 17:14
Hint $\rm\ \color{#C00}{g^d = 1},\,\ d\mid n\:\Rightarrow\: n = dk\:\Rightarrow\: g^n = (\color{#C00}{g^d})^k = \color{#C00}1^k = 1$