# Let $G$ a finite group of order $m$ with identity element $e$. Show that $g^m=e,\forall g\in G$

Let $$G$$ a finite group of order $$m$$ with identity element $$e$$.

1. Show that for each $$g\in G$$ exist a minimum $$k$$ such that $$g^k=e$$

2. Show that $$g^m=e,\forall g\in G$$ knowing that $$|G|=|N|\cdot|G/N|$$ holds for finite groups, where $$N$$ is any subgroup of $$G$$.

Part one: First notice that induced exponentiation from the group operation is a commutative operation, i.e. $$(gg)g=g(gg)$$ because any group operation is associative.

Now suppose that $$e\notin\{g,g^2,...,g^\ell\}$$, and suppose that $$g^{\ell+1}\neq e$$, then $$g^{\ell+1}\notin\{g,g^2,g^3,...,g^\ell\}$$. Proof: if $$g^{\ell+1}\in\{g,g^2,g^3,...,g^\ell\}$$ this mean that $$g^{\ell+1}=g^q=g^qg^{\ell+1-q}$$ and then $$g^{\ell+1-q}=e$$ what contradict the first assumption.

Then, using backward reasoning, if $$e\notin\{g,g^2,...,g^\ell\}$$ then $$g^q\neq g^p$$ for $$p\neq q\in\{1,...,\ell\}$$.

Now, because $$|G|=m$$ is finite exist some $$k\le m$$ such that $$g^k=e$$, i.e. $$\{g,g^2,...,g^k=e\}\subseteq G$$.

Part two: $$N_g=\{g,g^2,...,g^k=e\}$$ is a subgroup of $$G$$, i.e.

• $$e\in N_g$$
• if $$h\in N_g$$ then $$h^{-1}\in N_g$$, i.e. $$e=gg^{k-1}=gg^{-1}$$, $$e=g^2g^{k-2}=g^2(g^2)^{-1}$$, etc...
• for any $$h,j\in N_g$$ then $$hj\in N_g$$, i.e. if $$h=g^p$$ and $$j=g^q$$ then $$g^{p+q}\in N_g$$ (if $$p+q\le k$$ then this is obvious, and if $$p+q>k$$ then $$p+q=ak+b$$).

Because $$|N_g|=k$$ then we have that $$k$$ divides $$m$$ provided that $$|G|=|N_g|\cdot |G/N_g|$$, where $$G/N_g$$ is the set of equivalent classes defined by

$$a\sim b\iff a\in bN_g$$

Because $$m$$ is a multiple of $$k$$ then $$g^m=(g^k)^p=e^p=e$$, so we are done.

Questions:

• I have a bit of problem formalizing the "backward reasoning" of part one, I dont know how to write it more formally (using any kind of induction), can you let me some advice here?

Consider the sequence: $$g, g^2, g^3, \ldots, g^m, g^{m+1}$$ By closure, we know that each of these $m + 1$ elements belong to $G$. But since $|G| = m$, we know by the Pigeonhole Principle that some two elements in the sequence must match. That is, there exist $i, j \in \{1, \ldots, m + 1\}$ with $i < j$ such that: $$g^j = g^i$$ Multiplying both sides by $g^{-i}$, we get: $$g^{j-i} = e$$ and so we can take $k = j - i \in \{1, \ldots, m\}$.