Math 120A HW 7
Due Tuesday, December 2.
- 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.)
- List all abelian groups of order $108$ up to isomorphism.
- If $\mathbb{Z}_{108}$ does not appear in your list, explain which of the groups in your list it is isomorphic to, and why.
- Let $G$ be an abelian group of order $1024$ that is not isomorphic to $\mathbb{Z}_{1024}$. Prove that $G$ has a subgroup $H$ that is isomorphic to the Klein group $V$ (equivalently, isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$.) Hint: use the classification theorem.
- 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 $6$.)
- Calculate $\phi[4\mathbb{Z}]$, the image of $4\mathbb{Z}$ under $\phi$. (This will be a subgroup of $\mathbb{Z}_6$ because the image of a subgroup under a homomorphism is a subgroup.)
- Calculate $\phi^{-1}\big[\phi[4\mathbb{Z}]\big]$, the inverse image of your answer for part (a) under $\phi$.
Write your answer in the form $n\mathbb{Z}$. (Your answer will be a subgroup of $\mathbb{Z}$ because the inverse image of a subgroup under a homomorphism is a subgroup, and every subgroup of $\mathbb{Z}$ has the form $n \mathbb{Z}$ for some integer $n$.)
- Is $\phi^{-1}\big[\phi[4\mathbb{Z}]\big]$ equal to $4\mathbb{Z}$?
Note that "$\phi^{-1}[\ldots]$" is just a confusing notation
for the inverse image. There is no function $\phi^{-1}$ in this
problem because $\phi$ is not injective. So if you are writing things
like "$\phi^{-1}(\ldots)$" where "$\ldots$" is a group element, you are doing
something wrong.
-
Let $G$ and $H$ be groups and let $\phi: G \to H$ be a homomorphism
that is surjective. Prove that if $G$ is cyclic, then $H$ is cyclic.
Hint: If $G$ is cyclic, then it has some generator $g$. We want to show
that $H$ has a generator; what is the natural guess of a generator for $H$?
Does this natural guess work? How can you use the hypothesis that $\phi$
is surjective?
There are only 4 problems assigned this week
Optional practice problems from the book (these will not be graded): Section 11 Exercises 7, 26, 27; Section 13 Exercises 1, 2, 25, 27.