[Math] Elementary divisors of a group

abelian-groupsabstract-algebra

What are the elementary divisors of the group $\mathbb{Z}_{2} \oplus \mathbb{Z}_9 \oplus \mathbb{Z}_{35}$ and what are its invariant factors?
Do the same for $\mathbb{Z}_{26} \oplus \mathbb{Z}_{42} \oplus \mathbb{Z}_{49} \oplus \mathbb{Z}_{200} \oplus \mathbb{Z}_{1000}$.

So I found the elementary divisors of the first group to be {2,3,3,5,7}. I looked at two ways to write this set as {2,3,3,5,7} with invariant factor of (3,210), and set {2,9,5,7) with invariant factor (630).

I'm not sure about the other group. I found the elementary divisors to be {2,2,2,2,2,2,2,2,3,5,5,5,5,5,7,7,7,13}, or {$2^8,3,5^5,7^3,13$}. I'm just not sure what the invariant factors are. Can someone help me?

Best Answer

$\mathbb{Z}_2\oplus\mathbb{Z}_9\oplus\mathbb{Z}_{35}$

Elementary Divisors: $2,3^2,5,7$.

Invariant Factors: $630$.

$\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\oplus\mathbb{Z}_{49}\oplus\mathbb{Z}_{200}\oplus\mathbb{Z}_{1000}$

Elementary Divisors: $2,2,2^3,2^3,3,5^2,5^3,7,7^2,13$.

Invariant Factors: $2,2,1400,1911000$.

Elementary divisors are fairly self-explanatory. You break each $n$ in the $\mathbb{Z}_n$ pieces into its unique prime factorization, $n=p_1^{r_1}\cdots p_m^{r_m}$, where the $p_i$ are distinct primes and the $r_i >0$. We then rewrite $\mathbb{Z}_n$ as: $$\mathbb{Z}_n \cong \mathbb{Z}_{p_1^{r_1}}\oplus\mathbb{Z}_{p_2^{r_2}}\oplus\cdots\oplus\mathbb{Z}_{p_m^{r_m}}.$$ It is important to note that we cannot break the $\mathbb{Z}_9$ into $\mathbb{Z}_3\oplus\mathbb{Z}_3$ which is why the elementary divisors for the first group are $2,3^2,5,7$. For the invariant factors for the first group, see the example for the second group:

For the second group we see $$\mathbb{Z}_{26}\cong\mathbb{Z}_{13}\oplus\mathbb{Z}_2 \text{ and } \mathbb{Z}_{42}\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_3\oplus\mathbb{Z}_7$$ and so $$\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\cong\mathbb{Z}_{13}\oplus\mathbb{Z}_2\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_3\oplus\mathbb{Z}_7.$$ So in the expression $$\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\oplus\mathbb{Z}_{49}\oplus\mathbb{Z}_{200}\oplus\mathbb{Z}_{1000}$$ we can replace the $\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}$ with $$\mathbb{Z}_{13}\oplus\mathbb{Z}_2\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_3\oplus\mathbb{Z}_7.$$ The fact I have been using is that $$\mathbb{Z}_{mn}\cong\mathbb{Z}_m\oplus\mathbb{Z}_n$$ whenever $\gcd(m,n)=1$.

We continue in this way until we have rewritten $$H=\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\oplus\mathbb{Z}_{49}\oplus\mathbb{Z}_{200}\oplus\mathbb{Z}_{1000}$$ in the form $$\mathbb{Z}_{p_1^{r_1}}\oplus\mathbb{Z}_{p_2^{r_2}}\cdots\oplus\mathbb{Z}_{p_m^{r_m}}$$ where all the $p_i$ are primes and the $r_i>0$. Doing so, we get that $$H\cong \mathbb{Z}_{2}\oplus\mathbb{Z}_{13}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{3}\oplus\mathbb{Z}_{7}\oplus\mathbb{Z}_{7^2}\oplus\mathbb{Z}_{2^3}\oplus\mathbb{Z}_{5^2}\oplus\mathbb{Z}_{5^3}\oplus\mathbb{Z}_{2^3}.$$ For invariant factors, there is a trick once you know the elementary divisors. I will do the invariant factors for $\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\oplus\mathbb{Z}_{49}\oplus\mathbb{Z}_{200}\oplus\mathbb{Z}_{1000}$ now. I make this table out of the elementary divisors: $$\begin{array}{c|c|c|c|c} p=2 & p=3 & p=5 & p=7 & p=13 \\ \hline 2^3 & 3 & 5^3 & 7^2 & 13\\ 2^3 &1&5^2&7&1\\ 2&1&1&1&1\\ 2&1&1&1&1 \end{array}$$ The first row I label each of the primes that appear as elementary divisors. In the next row I write the highest power of the given prime in each column. In the next row I write the elementary divisors of the next powers that I haven't used. I put a $1$ in a column when I have already recorded all of the elementary divisors of a given prime. It may be helpful to write down the elementary divisors and record them in a table on paper and compare to see that you understand how to do this. Next, take the product of the numbers in each row, and this will yield your invariant factors. In this case we get the invariant factors are: $$2\cdot1\cdot1\cdot1\cdot1,2\cdot1\cdot1\cdot1\cdot1,2^3\cdot1\cdot5^2\cdot7\cdot1,2^3\cdot3\cdot5^3\cdot7^2\cdot13,$$ that is, the invariant factors are $$2,2,1400,1911000.$$

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