I am confused about this task:

Write $G= \mathbb {Z_1} \oplus \mathbb {Z_2} \oplus\mathbb {Z_3} \oplus\mathbb {Z_4} \oplus\mathbb {Z_5} \oplus\mathbb {Z_6} \oplus\mathbb {Z_7} \oplus\mathbb {Z_8} \oplus\mathbb {Z_9} \oplus\mathbb {Z_{10} } $ In Form of the main theorem about finite Abelian groups.

After my understanding it is right the definition of the theorem. Help would be very appreciated

  • $\begingroup$ What do you mean by $\mathbb{Z}_{i}$? In particular, what's $\mathbb{Z}_1$? $\endgroup$ – egreg Nov 22 '17 at 17:19
  • $\begingroup$ I am confused about the question. $\endgroup$ – user312648 Nov 22 '17 at 17:24
  • $\begingroup$ We defined it as $\mathbb{Z_1}=\{ 0 \}, \mathbb{Z_2}=\{ 0,1 \},$ and so on $\endgroup$ – wondering1123 Nov 22 '17 at 17:25
  • $\begingroup$ @cello I am confused too :( $\endgroup$ – wondering1123 Nov 22 '17 at 17:26
  • $\begingroup$ The fundamental theorem comes in two forms: one with the elementary divisors and the other with the invariant factors. Have you written down the elementary divisors? $\endgroup$ – Lozenges Nov 22 '17 at 17:58

Are you familiar with the Chinese Remainder Theorem?

This post will be useful reading: Why doesn't the Chinese remainder theorem contradict the Fundamental Theorem of Finitely Generated Abelian Groups?

Hint: In particular, what is true for $\mathbb{Z}/6\mathbb{Z}$ (but perhaps not true for $\mathbb{Z}/4\mathbb{Z}$)?

  • $\begingroup$ I do not see how this answers the question. Are you sure this answers the question? $\endgroup$ – user312648 Nov 22 '17 at 17:37
  • $\begingroup$ Certainly! What do you know about groups of order 4? How many of them are there? $\endgroup$ – bounceback Nov 22 '17 at 17:50

Recall that $\mathbb{Z}_m \oplus \mathbb{Z}_n \cong \mathbb{Z}_{mn}$ if and only if $m$ and $n$ are relatively prime. So you can just write each $i \in \{2, \ldots, 10\}$ as a product of powers of distinct primes (which are then relatively prime). Since $\mathbb{Z}_1$ is just the trivial group, you can exclude it from the direct sum.

\begin{align} G &\cong \mathbb {Z}_2 \oplus\mathbb {Z}_3 \oplus\mathbb {Z}_4 \oplus\mathbb {Z}_5 \oplus\mathbb {Z}_6 \oplus\mathbb {Z}_7 \oplus\mathbb {Z}_8 \oplus\mathbb {Z}_9 \oplus\mathbb {Z}_{10} \\ &\cong \mathbb {Z}_2 \oplus\mathbb {Z}_3 \oplus\mathbb {Z}_4 \oplus\mathbb {Z}_5 \oplus(\mathbb {Z}_2 \oplus \mathbb{Z}_3) \oplus\mathbb {Z}_7 \oplus\mathbb {Z}_8 \oplus\mathbb {Z}_9 \oplus(\mathbb {Z}_{2} \oplus \mathbb{Z}_5)\\ &\cong \mathbb {Z}_2 \oplus \mathbb {Z}_2 \oplus \mathbb {Z}_2 \oplus \mathbb {Z}_4 \oplus\mathbb {Z}_8 \oplus\mathbb {Z}_3 \oplus \mathbb {Z}_3 \oplus \mathbb{Z}_9 \oplus\mathbb {Z}_5 \oplus\mathbb {Z}_5 \oplus\mathbb {Z}_7 \\ \end{align}

You can now read the elementary divisors of $G$: $$2,2,2,4,8,3,3,9,5,5,7$$

To find the invariant factors, write the powers of primes as columns of a table in descending order.

$$\left[\begin{array}{cccc|c} 8 & 9 & 5 & 7 & 2520 \\ 4 & 3 & 5 & 1 & 60\\ 2 & 3 & 1 & 1 & 6\\ 2 & 1 & 1 & 1 & 2\\ 2 & 1 & 1 & 1 &2 \end{array}\right]$$

The invariant factors are products of elements in every row: $2520, 60, 6, 2, 2$.

Hence $$G \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_{60} \oplus \mathbb{Z}_{2520}$$


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