I'm trying to follow this proof in my textbook, Contemporary Abstract Algebra by Gallian (p231) but I'm having trouble understanding what's going on. He writes

Let G be a finite Abelian group of order $p^nm$, where p is a prime that does not divide m. Then $G=H\times K$ where $H=\{ x\in G | x^{p^n}=e\}$ and $K=\{ x\in G | x^m=e\}.$ Moreover, $|H|=p^n$.

proof: It is an easy exercise to prove that H and K are subgroups of G. Because G is Abelian, to prove that $G=H\times K$ we need only to prove that $G=HK$ and $H\cap K = \{ e\}$. Since we have $gcd(m,p^n)=1$, there are integers $s$ and $t$ such that $1=sm+tp^n$.

So far, so good.

For any $x\in G$, we have $x=x^1=x^{sm+tp^n}=x^{sm}x^{tp^n}$, and by Corollary 4 of Lagrange's theorem (Theorem 7.1), $x^{sm}\in H$ and $x^{tp^n}\in K$.

And he goes on to finish the proof.

My problem is that the corollary he is referring to says

Let $G$ be a finite group and let $a\in G$. Then $a^{|G|}=e$.

And I can't figure out how he uses this to get from



$x^{sm}\in H$ and $x^{tp^n}\in K$.

What's going on here?

  • $\begingroup$ $(x^{sm})^{p^n}=e$ would mean $x^{sm} \in H$ $\endgroup$ – vnd Feb 22 '16 at 18:43
  • $\begingroup$ @vnd I guess I don't understand how $x=x^{sm}x^{tp^n}$ shows that $(x^{sm})^{p^n}=e$ $\endgroup$ – Zachary F Feb 22 '16 at 19:00
  • $\begingroup$ Just note that $(x^{sm})^{p^n}=(x^{s})^{mp^n}$ and that $|G|=mp^n$. $\endgroup$ – vnd Feb 22 '16 at 19:02
  • $\begingroup$ @vnd But I don't have $(x^{sm})^{p^n}$, I have $x^{sm+tp^n}$ $\endgroup$ – Zachary F Feb 22 '16 at 19:09
  • $\begingroup$ The point is that $x^{sm}$ is an element of $G$ such that when it is raised to the power of $p^n$, gives the identity. Now this is the property an element of $G$ should possess to be a member of $H$. $\endgroup$ – vnd Feb 22 '16 at 19:11

You simply have $(x^sm)^{p^n}=(x^{mp^n})^s=e^s=e$ by the corollary, hence by definition $x^{sm}$ is in $H$.

Similarly $x^{tp^n}$ lies in $K$.


Since $x \in G$, then $x^{p^nm}=e.$ Hence $x^{sp^nm}=e^s=e.$


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