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Find the primary decomposition and invariant factor decomposition of $\mathbb{Z}/180\mathbb{Z}$.

For the invariant factor decomposition, we just find the prime factorization of $180$ and write $180=5\times3\times3\times2\times2$. So $\mathbb{Z}/180\mathbb{Z} = \mathbb{Z}/6\mathbb{Z} \oplus \mathbb{Z}/30\mathbb{Z}$, right? But what if we said, for example, $\mathbb{Z}/180\mathbb{Z} = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/90\mathbb{Z}$ or $\mathbb{Z}/180\mathbb{Z}= \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/60\mathbb{Z}$. Would those aslo count as an invariant factor decomposition (since we know that $2|90$ and $3|60$)?

For the primary factor decomposition: Our textbook says...If $G$ is any abelian group, then its $p$-primary component is $G_p = \{ a\in G: p^na=0$ for some $n \geq 1\}$.

Every finite abelian group $G$ is the direct sum of its p-primary components:

$G = G_{p_1} \bigoplus ... \bigoplus G_{p_n}$

For this question, we know that the primes are 2, 3, and 5. So the primary decomposition is $G = G_2 \bigoplus G_3 \bigoplus G_5$. Do you think that is correct?

Thanks in advance

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Thank you. I also have a question from this link: For the fourth example, it says that the primary decomposition is $Z_2 \bigoplus Z_4 \bigoplus Z_8 \bigoplus Z_3 \bigoplus Z_3 \bigoplus Z_5 \bigoplus Z_5$. But we know that 4 and 8 are not prime. Also, why do we have to repeat 3 and 5 two times? – user58289 Jan 19 '13 at 1:48
Thanks a lot. So for the original question that I asked, do I have to say $G_p = Z_2 \bigoplus Z_2 \bigoplus Z_3 \bigoplus Z_3 \bigoplus Z_5$? Also, for the question in the link, why do we write $Z_{24} = Z_3 \bigoplus Z_8$ instead of writing $Z_{24} = Z_3 \bigoplus Z_2 \bigoplus Z_2 \bigoplus Z_2$? Is it because 8 and 3 are relatively prime? So is writing $Z_{24} = Z_3 \bigoplus Z_2 \bigoplus Z_2 \bigoplus Z_2$ wrong? – user58289 Jan 19 '13 at 13:33
up vote 4 down vote accepted

$\mathbb Z_{180}\simeq \mathbb Z_4\oplus\mathbb Z_9\oplus\mathbb Z_5$ (here I've used repeatedly that $\mathbb Z_{mn}\simeq\mathbb Z_m\oplus\mathbb Z_n$ whenever $(m,n)=1$). This shows that for this group the $2$-primary component is $\mathbb Z_4$, the $3$-primary component is $\mathbb Z_9$, and the $5$-primary component is $\mathbb Z_5$.

For the invariant factor decomposition: $180$ is the only invariant factor of this group, since the elementary divisors are $2^2$, $3^2$, and $5$. It can't be true that $\mathbb Z_{180}\simeq\mathbb Z_6\oplus\mathbb Z_{30}$: if this holds, then $30$ will kill every element of $\mathbb Z_{180}$, which is obviously false. The same can be said for the other decompositions.

Edit. The OP asked in the comments about similar decompositions for the group $\mathbb Z_{24}\oplus\mathbb Z_{30}\oplus\mathbb Z_{20}.$ In this case the elementary divisors are $2^3$, $3$, $2$, $3$, $5$, $2^2$, $5$. This shows that the $2$-primary component of this group is $\mathbb Z_8\oplus\mathbb Z_2\oplus\mathbb Z_4$, the $3$-primary component is $\mathbb Z_3\oplus\mathbb Z_3$, and the $5$-primary component is $\mathbb Z_5\oplus\mathbb Z_5$.

The invariant factor decomposition can be found by using the elementary divisors: the greatest invariant factor is $120=2^3\cdot 3\cdot 5$, the next one is $60=2^2\cdot 3\cdot 5$, and the last one is $2$. Thus we have the following invariant factor decomposition: $\mathbb Z_{2}\oplus\mathbb Z_{60}\oplus\mathbb Z_{120}.$

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