Can you check my proof and let me some advice about formal writing please?

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.


  • 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?

  • Any other advice is welcome. Thank you in advance!

  • 1
    $\begingroup$ Use Lagrange's theorem. It is almost immediate. $\endgroup$
    – DonAntonio
    Sep 18, 2016 at 19:22
  • $\begingroup$ @DonAntonio in the book that Im reading this exercise is before to Lagrange theorem (anyway I dont know this theorem by now). $\endgroup$
    – Masacroso
    Sep 18, 2016 at 19:24
  • $\begingroup$ Ok. I am so fixed into using that theorem that right now, from the top of my head, I can't see how to solve this problem without Lagrange's theorem or something equivalent to it. $\endgroup$
    – DonAntonio
    Sep 18, 2016 at 19:25
  • $\begingroup$ Your argument is essentially correct, if not elegantly written. You've actually almost proved Lagrange's Theorem. $\endgroup$ Sep 18, 2016 at 19:44
  • $\begingroup$ May I ask from which textbook this exercise comes? $\endgroup$
    – Akira
    Mar 18, 2019 at 8:22

1 Answer 1


Here's an alternate proof for part 1.

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 $k = j - i \in \{1, \ldots, m\}$.


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