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Up to isomorphic, how many additive abelian groups $G$ of order 16 have the property such that $x+x+x+x=0$, for each $x$ in G?

My question is that which theorem can be used? My answer is 3. Is that right?

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en.wikipedia.org/wiki/… –  blitzer Oct 25 '12 at 17:44

3 Answers 3

up vote 6 down vote accepted

It follows from the hypothesis (and the Structure Theorem for Abelian Groups) that such groups will be products of $\Bbb Z_2$ and/or $\Bbb Z_4$. In particular, it will be one of $\Bbb Z_4\times\Bbb Z_4$, $\Bbb Z_4\times\Bbb Z_2\times\Bbb Z_2$, or $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$. Your answer is correct.

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Do you know the classification theorem for finitely generated abelian groups? It states that $$G \simeq \mathbb{Z}^j \oplus \mathbb{Z}_{p_1^{r_1}}\oplus \mathbb{Z}_{p_2^{r_2}}\oplus ... \oplus \mathbb{Z}_{p_k^{r_k}}$$ where $k$ and the $r_i$'s are positive and the $p_i$'s are non necessarily distinct primes.

Thus, there are only a finite number of abelian groups with order 16, and only a subset of these satisfy your condition, which is that every element has order less than or equal to 4.

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Using the big gun "classification of finite abelian groups", we see that there are $(\mathbb Z/4\mathbb Z)^2$, $\mathbb Z/4\mathbb Z\times (\mathbb Z/2\mathbb Z)^2$ and $(\mathbb Z/2\mathbb Z)^4$.

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