Is the cyclic group $\langle x\rangle$ always a subgroup of $G$ for any $x\in G$? I have been thinking about the following:

If $G$ is a finite group and $x\in G$ an element of order $n$ is then $\langle x\rangle$ always a subgroup of $G$?

I have the definition that $\langle x\rangle=\{1,x,x^2,...,x^{{\rm ord}(x)-1}\}$
Do check this I know that I should check the following three points:

*

*$1_G\in \langle x\rangle$. This is clear since $x^0=1$.

*Stable under inverses: Let $u\in \langle x\rangle$. Then $u=x^i$ for some $1\leq i \leq n$. I claim that $u^{-1}=(x^{n-1})^i$ is the inverse of $u$. Indeed $$x^i(x^{n-1})^i=x^i(x^{ni-i})=x^{ni}=(x^n)^i=1_G^i=1_G$$ And since $\langle x\rangle$ is abelian $u^{-1}$ is indeed the inverse of $u$. Furthermore  it is clear that $u^{-1}\in \langle x\rangle$.

*Stable under composition: let $u,v\in \langle x\rangle$, then $u=x^i, v=x^j$ for some $1\leq i,j\leq n$. Then $uv=x^ix^j=x^{i+j}\in \langle x\rangle$
According to this proof I think my claim should be correct but I'm not 100% sure, thats why I wanted to ask if someone could take a look. Thanks a lot.
 A: Yes.
In fact, neither the order of the element $x$ nor of the group $G$ need be finite. I will concentrate on the case when ${\rm ord}\, x$ is finite.
Use the one-step subgroup test.
Since $e=x^0\in\langle x\rangle$, we have $\langle x\rangle\neq \varnothing$.
Let $y\in\langle x\rangle$. Then $y=x^n$ for some $n\in\Bbb Z$. But all powers of $x$ are in $G$ by closure of $G$. Hence $y\in G$. Hence $\langle x\rangle\subseteq G$.
Let $a=x^r, b=x^s\in \langle x\rangle$. Then
$$\begin{align}
ab^{-1}&=x^r(x^s)^{-1}\\
&=x^rx^{-s}\\
&=x^{r-s},
\end{align}$$
which is in $\langle x\rangle$ by taking the power modulo ${\rm ord}\, x$: we can do this by Euclid's algorithm, by finding $0\le m<{\rm ord}\, x$ and $q\in\Bbb Z$ such that $r-s=q{\rm ord}\, x+m$, since then
$$\begin{align}
x^{r-s}&=x^{q{\rm ord}\, x+m}\\
&=(x^{{\rm ord }\, x})^qx^m\\
&=e^qx^m\\
&=x^m.
\end{align}$$
Hence $ab^{-1}\in\langle x\rangle$.
Hence $\langle x\rangle\le G$.
A: We have disscussed several solutions, and as suggested from @user1729 I put in a correct solution with the comments. I use the way I started in the question. So we check the following three points:

*

*$1_G\in \langle x\rangle$. This is clear since $x^0=1$.


*Stable under inverses: Let $u\in \langle x\rangle$. Then $u=x^i$ for some $1\leq i \leq n$. I claim that $u^{-1}=(x^{n-1})^i$ is the inverse of $u$. Indeed $$x^i(x^{n-1})^i=x^i(x^{ni-i})=x^{ni}=(x^n)^i=1_G^i=1_G$$ And since $\langle x\rangle$ is abelian $u^{-1}$ is indeed the inverse of $u$. Furthermore  it is clear that $u^{-1}\in \langle x\rangle$.


*Stable under composition: let $u,v\in \langle x\rangle$, then $u=x^i, v=x^j$ for some $1\leq i,j\leq n$. Then $uv=x^ix^j=x^{i+j}$. Here it is not clear that $x^{i+j} \in \langle x\rangle$. But remark that since $0\leq i,j<n$ implies that $i+j<2n$ thus we can find $0\leq k<n$ such that $i+j=n+k$ and therefore $$x^{i+j}=x^{n+k}=x^nx^k=x^k\in \langle x \rangle$$
I hope this works now
A: Well it is correct . Hence it is not necessary that $x$ is an element of finite order neither should $G$ be finite .
I propose the following proof .
let $x$ be an element of  $G$ (with no condition on $x$)
Consider the application, $f: \mathbb{Z} \to G$
, such that for all k in $\mathbb{Z}$ , $f(k)=x^k$
You can proof that $f$ is a morphism of groups .
Note that :  any two integers $k,l$ ;  $f(k+l)=f(k)∗f(l)$
Then $f(\mathbb{Z})=\{x^k\mid k\in\mathbb{Z}\}$ is a subgroup of G. i.e $  \langle x\rangle$  is a subgroup of G.
