Let $G$ be a finite group. I want to prove that if $g\in G$ then $\langle g\rangle$ is a subgroup of $G$. I use the subgroup test and show it is nonempty, closed under the operation and the inverses are closed. If $g\in G$ then $g\in \langle g\rangle$ obviously. If $h,k \in G$ then $h=g^m$, $k=g^n$ for integers $m$ and $n$. Hence $hk=g^{n+m}\in \langle g\rangle$. Lastly, if $h \in \langle g\rangle$ then $h=g^n$ for some integer $n$ and then $h^{-1}=g^{-n}\in\langle g\rangle$. However, I am a bit bewildered since I haven't used the fact that $G$ is a group?
If $g \in G$ then $\langle g\rangle$ is a subgroup of G
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group-theory
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1Would you mind telling us what you mean with the "subgroup test"? – 2017-02-09
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0@Wojowu it's implicit in the question: $H$ is a subgroup if it is non-empty, closed under "the operation" and closed under inverses. – 2017-02-09
2 Answers
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Your proof is perfectly correct. You are in fact implicitly using the associativity of the group product when you write things like $g^mg^n = g^{m+n}$. In particular: associativity allows us to refer to, for example, $$ g^3 = (gg)g = g(gg) $$ without any ambiguity.
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I assume that you define $\langle g\rangle=\{g^n:n\in\mathbb{Z}\}$.
It is a subgroup.
$1=g^0\in\langle g\rangle$
If $x=g^m,y=g^n\in\langle g\rangle$, then $xy=g^{m+n}\in\langle g\rangle$
If $x=g^m\in\langle g\rangle$, then $x^{-1}=g^{-m}\in\langle g\rangle$.
Where did we use that $G$ is a group? For instance in part 2, where $g^mg^n=g^{m+n}$ is used. Similarly it is used in part 3 and part 1.
Note that $a^{m}a^n=a^{m+n}$ is valid in every monoid, for $m,n\ge0$; one can extend this to all integers $m$ and $n$ only by assuming the element $a$ has an inverse.
Actually, if $X$ is a monoid and $x\in M$ is invertible, then $\{x^n:n\in\mathbb{Z}\}$ is a subgroup of $M$, for the same reasons.