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?
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?
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.
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 $(\mathbb Z/2\mathbb Z)^4$.
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.