Let $G$ be a group and let $g\in G$ be an element of finite order n.

(i) For $m\in \Bbb Z$, $g^m=g^r$ where $r$ is the remainder on division of $m$ by $n$.

(ii)The order of the cyclic subgroup $\langle g\rangle$ generated by $g$ is $n$.

Proof of (ii):

By (i), every power $g^m$ of $g$ is equal to one of the $n$ elements $e,g,...,g^{n-1}$.No two of these elements are equal, for if $0\le j<i\le n-1$ are such that $g^i=g^j$ then $0<i-j<n$ and $g^{i-j}=e$, contradicting the definition of the order of $g$. Thus, $\langle g\rangle={e,g,...,g^{n-1}}$ has order $n$.

My problem is I literally don't understand this proof. First statement is bit cryptic for me what does "every power $g^m$" of $g$" actually means? Does it mean $g^{m+1}$ or $g$ with power $m$? Well if it is $g$ with power $m$ then this statement is might be true since $0\le r<n$ so there might exist $g^r$ such that $g^m$ equal to it . Second statement says that it is contradicting the definition of the order of $g$ and I don't know how the hell does it do contradiction I really don't get it. Now third statement is just a result which everyone knows.

(I know this question might sound dumb but I really don't have ability to understand it myself anyway thanks for taking time to read it I hope you can explain. Good Luck!)

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    $\begingroup$ Every power $g^m$ means $\{g^m:m \in \mathbb{Z}\}.$ Second statement contradicts the definition becomes $i-j<n,$ and $g^{i-j}=e,$ which means order of $g \leq i-j <n.$ $\endgroup$ Jun 10, 2020 at 10:11

2 Answers 2


If $|g|=n$, then:

\begin{alignat}{1} \langle g \rangle &:= \{g^m, m\in \mathbb{Z}\} \\ &= \{g^{kn+r}, k\in \mathbb{Z}, r=0,\dots,n-1\} \\ &= \{g^r, r=0,\dots,n-1\} \\ \end{alignat}

Now, we have to prove that the $n$ elements $g^r, r=0,\dots,n-1$, are all distinct, so as to conclude that $|\langle g \rangle|=n=|g|$. Seeking for a contradiction, let's suppose they are not all distinct; then, $\exists i,j, \space0\le i< j\le n-1$, such that $g^i=g^j$; but then, being $0<j-i\le n-1$, we have a positive integer strictly smaller than $n$, $l:=j-i$, such that $g^l=e$, in contradiction with the hypothesis that $g$ has order $n$. Therefore $\langle g \rangle$ has exactly $n$ (distinct) elements, i.e. $|\langle g \rangle|=n=|g|$.

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    $\begingroup$ $g^i=g^j \Rightarrow g^{j-i}=e$, not $i=j$; on the contrary, remember that you are assuming $i<j$, with both $i$ and $j$ in between $0$ and $n-1$ (both included). $\endgroup$
    – user750041
    Jun 11, 2020 at 6:33
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    $\begingroup$ How would you else take two distinct elements $g^i, g^j$ of the collection $\{g^r, r=0,\dots,n-1\}$, if not assuming $i\ne j$? $\endgroup$
    – user750041
    Jun 11, 2020 at 6:53
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    $\begingroup$ Basically you have a collection (with possibly replicated elements), and you want to prove it is indeed a set (no replicated elements). $\endgroup$
    – user750041
    Jun 11, 2020 at 6:56
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    $\begingroup$ Right, a contradiction. So $i\ne j$, with $i,j\in \{0,\dots,n-1\}$, implies $g^i\ne g^j$, and indeed $\langle g \rangle$ is made of $n=|g|$ distinct elements (its order). $\endgroup$
    – user750041
    Jun 11, 2020 at 7:02
  • $\begingroup$ Thanks! Now I understand that it is $g^l\neq e $ since $g^n$ or $g^0$ should be the one which should be $e$. $\endgroup$
    – Not Tfue
    Jun 11, 2020 at 7:57

Sometimes the best way to understand and abstract proof is to do a concrete example.

Let's take $G = S_5$ the group of all permutations of 5 symbols $A,B,C,D,E$. So for example $(A\,B)(C\,D)$ is the permutation that swaps $A$ and $B$, swaps $C$ and $D$ and leaves $E$ alone.

Let's take $g = (A\,C\,E)$, which has order 3.

Now lets look at the powers of g:

  • $g^0 = ()$ the identity
  • $g^1 = (A\,C\,E)$ the identity
  • $g^2 = (A\,E\,C)$ the identity
  • $g^3 = ()$ the identity
  • $g^4 = (A\,C\,E)$ the identity
  • $g^5 = (A\,E\,C)$ the identity
  • $g^6 = ()$ the identity

you can see the repeating happening. This is what is being described as every power of g is one of 3 different things.

Now $\langle g \rangle$ is the group generated by $g$. As a set it contains those 3 different powers of $g$, so $\langle g \rangle = \{(), (A\,C\,E), (A\,E\,C)\}$. As a group you can multiply them together and the result will always still be one of those 3 things.

So we have seen that $\langle g \rangle$ is a subgroup of $S_5$ and it has order 3.


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