Math 120A HW 7 Solutions
- For this problem, you can use the classification theorem for finite abelian
groups even though we did not prove it. It can be found in the textbook
as Theorem 11.12. (Note that the book talks more generally about finitely
generated abelian groups, which we have not defined. The factors
"${}\times \mathbb{Z} \times \mathbb{Z} \times \cdots \times \mathbb{Z}$" at the
end do not show up when dealing with finite groups.)
- There are six:
- $\mathbb{Z}_{2^2} \times \mathbb{Z}_{3^3}$,
- $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_{3^3}$,
- $\mathbb{Z}_{2^2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_3$,
- $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_{3^2} \times \mathbb{Z}_3$,
- $\mathbb{Z}_{2^2} \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$ and
- $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$.
- $\mathbb{Z}_{2^2} \times \mathbb{Z}_{3^3}$ is cyclic because $2^2$ and $3^3$ are relatively prime. So because it has order $108$, it is isomorphic to $\mathbb{Z}_{108}$.
- Because $1024$ is a power of $2$,
by the classification theorem for finite abelian groups
$G$ is isomorphic to $\mathbb{Z}_{2^{r_1}} \times \cdots \times
\mathbb{Z}_{2^{r_k}}$ for some exponents ${r_1}, \ldots,
{r_k}$.
Without loss of generality we may assume $G =
\mathbb{Z}_{2^{r_1}} \times \cdots \times \mathbb{Z}_{2^{r_k}}$.
Because $G$ is not isomorphic
to $\mathbb{Z}_{1024}$ there is more than one factor, so $k \ge 2$.
The direct product of subgroups
$\{0,2^{r_1-1}\} \times \{0,2^{r_2-1}\} \times \{0\} \times \cdots \times \{0\}$,
which is a subgroup of the direct product $G$, is isomorphic to
$\mathbb{Z}_2 \times \mathbb{Z}_2$ because its two nontrivial factors are each isomorphic to $\mathbb{Z}_2$.
- Consider the homomorphism $\phi : \mathbb{Z} \to \mathbb{Z}_6$ defined by $\phi(a) = a \mathbin{\text{mod}} 6$ (the remainder when $a$ is divided by $6$.)
- $\phi[4\mathbb{Z}] = \{0,2,4\}$.
- $\phi^{-1}\big[\phi[4\mathbb{Z}]\big] = \phi^{-1}[\{0,2,4\}] = 2\mathbb{Z}$.
- $\phi^{-1}\big[\phi[4\mathbb{Z}]\big] \ne 4\mathbb{Z}$.
-
Fix a generator $g$ of $G$. We claim that $\phi(g)$ is a generator of $H$.
Let $b \in H$.
Because $\phi$ is surjective, $b = \phi(a)$ for some $a \in G$.
Because $g$ generates $G$, we have $a = g^n$ for some integer $n$.
Then $b = \phi(g^n) = \phi(g)^n$ because $\phi$ is a homomorphism.