# Explaination of proof of Property of cyclic subgroup.

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 $$0\le j are such that $$g^i=g^j$$ then $$0 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 $$0\le r so there might exist $$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!)

• 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 $g \leq i-j <n.$ Jun 10, 2020 at 10:11

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 $$0, we have a positive integer strictly smaller than $$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|$$.

• $g^i=g^j \Rightarrow g^{j-i}=e$, 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).
– user750041
Jun 11, 2020 at 6:33
• 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 $i\ne j$?
– user750041
Jun 11, 2020 at 6:53
• Basically you have a collection (with possibly replicated elements), and you want to prove it is indeed a set (no replicated elements).
– user750041
Jun 11, 2020 at 6:56
• 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).
– user750041
Jun 11, 2020 at 7:02
• Thanks! Now I understand that it is $g^l\neq e$ since $g^n$ or $g^0$ should be the one which should be $e$. 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.