The purpose of this post is to show how a finite direct sum of finite cyclic groups

$\Large \Bbb Z_{m_1} \oplus \Bbb Z_{m_2} \oplus \dots \oplus \Bbb Z_{m_n}$

can be rearranged so that their orders are in increasing divisional form: $m_1|m_2|\dots | m_n$.

We use the fact that if $p, q$ are coprime, then $\large \Bbb Z_p \oplus \Bbb Z_q = \Bbb Z_{pq}$.

(We'll use equality $=$ for isomorphism $\cong$ of groups.)

Let $p_1, p_2, \dots p_k$ be the list of prime numbers in the prime factorizations of all the integers $m_1, \dots, m_n$.

Write each $m_j$ in its prime power factorization $\large m_j = p_1^{a_{j1}}p_2^{a_{j2}} \dots p_k^{a_{jk}}$. Therefore

$\Large \Bbb Z_{m_j} = \Bbb Z_{p_1^{a_{j1}}} \oplus \Bbb Z_{p_2^{a_{j2}}} \oplus \dots \oplus \Bbb Z_{p_k^{a_{jk}}}$

and so the above direct sum $\large \Bbb Z_{m_1} \oplus \Bbb Z_{m_2} \oplus \dots \oplus \Bbb Z_{m_n}$ can be written out in matrix/row form as the direct sum of the following rows:

$\Large\Bbb Z_{p_1^{a_{11}}} \oplus \Bbb Z_{p_2^{a_{12}}} \oplus \dots \oplus \Bbb Z_{p_k^{a_{1k}}}$

$\Large\Bbb Z_{p_1^{a_{21}}} \oplus \Bbb Z_{p_2^{a_{22}}} \oplus \dots \oplus \Bbb Z_{p_k^{a_{2k}}}$

$\Large \vdots$

$\Large\Bbb Z_{p_1^{a_{n1}}} \oplus \Bbb Z_{p_2^{a_{n2}}} \oplus \dots \oplus \Bbb Z_{p_k^{a_{nk}}}$

Here, look at the powers of $p_1$ in the first column. They can be permuted / arranged so that their powers are in increasing order. The same with the powers of $p_2$ and the other $p_j$, arrange their groups so that the powers are increasing order. So we get the above direct sum isomorphic to

$\Large\Bbb Z_{p_1^{b_{11}}} \oplus \Bbb Z_{p_2^{b_{12}}} \oplus \dots \oplus \Bbb Z_{p_k^{b_{1k}}}$

$\Large\Bbb Z_{p_1^{b_{21}}} \oplus \Bbb Z_{p_2^{b_{22}}} \oplus \dots \oplus \Bbb Z_{p_k^{b_{2k}}}$

$\Large \vdots$

$\Large\Bbb Z_{p_1^{b_{n1}}} \oplus \Bbb Z_{p_2^{b_{n2}}} \oplus \dots \oplus \Bbb Z_{p_k^{b_{nk}}}$

where, for example, the exponents $b_{11} \le b_{21} \le \dots \le b_{n1}$ are a rearrangement of the numbers $a_{11}, a_{21}, \dots, a_{n1}$ (in the first column) in increasing order. Do the same for the other columns.

Now put together each of these rows into cyclic groups by multiplying their orders, thus

$\Large\ \ \Bbb Z_{N_1}$

$\Large \oplus \Bbb Z_{N_2}$

$\Large \vdots$

$\Large \oplus \Bbb Z_{N_n}$

where

$\large N_1 = p_1^{b_{11}} p_2^{b_{12}} \dots p_k^{b_{1k}}$,

$\large N_2 = p_1^{b_{21}} p_2^{b_{22}} \dots p_k^{b_{2k}}$,

$\large \vdots$

$\large N_n = p_1^{b_{n1}} p_2^{b_{n2}} \dots p_k^{b_{nk}}$.

In view of the fact that the $b_{1j} \le b_{2j} \le \dots \le b_{nj}$ is increasing for each $j$, we see that $N_1 | N_2 | \dots | N_n$, as required. $\blacksquare$

## No comments:

## Post a Comment