1
$\begingroup$

Suppose a finite abelian group $G$ has invariant factors $(m_1, m_2, ..., m_k)$. Show that $G$ has an element of order $s$ if and only if $s$ divides $m_k$.

Because these are the invariant factors, we then know that $G \cong \mathbb{Z}_{m_1} \times \mathbb{Z}_{m_2} ... \times \mathbb{Z}_{m_k}$. Also, because it is in invariant factor form, $m_1$ divides $m_2$ ..., $m_{k-1}$ divides $m_k$. I'm not sure how to proceed. I know that the order of $s$ must divide the order of $|G|=m_1*m_2*...*m_k$, but I'm not sure how to argue that it must divide $m_k$ specifically.

$\endgroup$
0
$\begingroup$

Let $x$ be an element of order $s$.
Write $$x=(x_1,x_2,\dots,x_k)$$ where $x_i\in \Bbb{Z}_{m_k}$ Then $$s=\text{lcm }(|x_1|,|x_2|,\dots,|x_k|) $$ Note that $|x_i|$ divides $m_i$ divides $m_k$.
Hence we conclude that $s$ divides $m_k$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.