3

I want to prove the following statement:

Let $G$ be a finite abelian group, let $m = \max\{|g|:g\in G\}$, and let $H$ be a cyclic subgroup of $G$ such that $|H|=m$. If $a\in G $ and $m'=|Ha|$, then $Ha$ has an element of order $m'$.

It is clear that the function $h\mapsto ha$ from $H$ to $Ha$ is bijective. Hence, $$m'=|Ha| = |H| = m.$$ Therefore, our aim is to prove that there exists an element in $Ha$ of order $m$.

We know that $H = \langle h\rangle$ for some $h\in G$, and $|h|=m$. Then, $$Ha = \{h^ka:0\leq k\leq m-1\}.$$ However, I have not been able to make the correct choice of an integer $k$ between $0$ and $m-1$ to get the element that we are looking for.

Edit: I know that $g^m=e$ for all $g\in G$, since $|g|$ is a factor of $m$. This fact was proved here. I was trying to apply this property, but it was useless.

Shaun
  • 44,997
Lord Vader
  • 961

1 Answers1

1

Let $a$ have order $l$. Let the prime power decomposition of $m$ be $P_1...P_uQ_1...Q_v$ where the $Q_i$ divide $l$ and the $P_i$ do not divide $l$. Let $Q$ be the product of the $Q_i$.

Consider $g=h^Qa$ and let $m=rn$ where $r$ is prime. Then $g^n=h^{Qn}a^n$.

If $r$ divides a $P_i$. Then $a^n$ is the identity and $h^{Qn}$ has order $r$. Hence $P_i$ divides the order of $g$.

If $r$ divides $Q_i$. Then $h^{Qn}$ is the identity and $a^n$ has order $r$. Hence $Q_i$ divides the order of $g$.

Thus $m$ divides the order of $g$, as required.

  • I don't understand your argument. First of all, in the description of the post I proved that $m'=m$, so there are no $P_i$'s. It follows that $Q=Q_1\ldots Q_v=m$. Since the order of $h$ is $m$, we have that $g = h^ma=a$. Therefore, you are saying that the order of $a$ is $m$, but this is not necessarily true because $a$ is an arbitrary element of $G$. – Lord Vader Feb 01 '22 at 17:37
  • I was using $m'$ for the order of $a$ - I've changed that now. –  Feb 01 '22 at 17:54
  • Why do you say that if $r\mid P_i$ then $a^n$ is the identity and $|h^{Qn}|=r$? – Lord Vader Feb 01 '22 at 17:59
  • The order of $a$ does not include the highest power of $r$ that divides $n$ and so $n/r$ suffices to make it the identity. –  Feb 01 '22 at 18:05
  • On the other hand, $h$ does require the highest power of $r$. –  Feb 01 '22 at 18:06
  • Why does $r$ divide $n$? – Lord Vader Feb 01 '22 at 18:24
  • Sorry to confuse you again. Coming back to this question cold I muddled up the letters in my comment. In my previous comment replace n by m. So m/r suffices and this is just n. –  Feb 01 '22 at 18:28