# Having trouble determining the quotient group in an algebraic topology course

I am working on an algebraic topology course (hatcher's book) and it has been quite a time since I took akgebra. I have my exams soon and I want a suggestion for a chapter or a resource online that helps me understand this:

For example, if we take the following answer, I understand everything related to the geometry and topology of the answer but when it comes to the last step, I am struggling with algebra:

https://math.stackexchange.com/a/58844/752801

How can i see that $$\mathbb{Z} \oplus \mathbb{Z} / \langle 2\mathbb{Z}(1,1)\rangle= \mathbb{Z} \oplus\mathbb{Z_2}?$$

I would be grateful if someone can explain this answer but also if someone can give me a resource that helps me understand this type of quotient group specifically for this course.

Maybe the clearest way to see that $$(\mathbb Z \oplus \mathbb Z) / \langle (2, 2) \rangle = \mathbb Z \oplus \mathbb 2 \mathbb Z$$ is to perform a change of basis on $$\mathbb Z \oplus \mathbb Z$$.

Usually when we think of $$\mathbb Z \oplus \mathbb Z$$, we think in terms of the standard basis, $$\{(1, 0), (0, 1)\}$$. But there are other bases for $$\mathbb Z \oplus \mathbb Z$$!

One such alternative basis is $$\mathcal B = \{(1, 1), (0, 1) \}$$. Now $$(2, 2)$$ can be expanded in terms of this alternative basis as $$(2, 2) = 2(1,1) + 0(0, 1)$$. In other words $$(2,2) = (2, 0)_{\mathcal B}$$ (if that notation makes sense).

Thus $$(\mathbb Z \oplus \mathbb Z) / \langle (2, 2) \rangle$$ is isomorphic to $$(\mathbb Z \oplus \mathbb Z) / \langle (2, 0) \rangle$$ (via the above change of basis), and obviously $$(\mathbb Z \oplus \mathbb Z) / \langle ( 2, 0) \rangle$$ is isomorphic to $$\mathbb Z_2 \oplus \mathbb Z$$.

(By the way, there is a typo in your post - it's not isomorphic to $$\mathbb Z \oplus 2 \mathbb Z$$!)

This begs the question:

Given a subgroup $$G$$ of $$\mathbb Z^{\oplus n}$$ (defined by a set of generators), is there a systematic algorithm that will find an alternative basis $$\mathcal B = \{ \mathbf v_1, \dots, \mathbf v_n \}$$ for $$\mathbb Z^{\oplus n}$$, a $$k \in \{0, \dots, n\}$$ and integers $$c_1, \dots, c_k$$ such that $$\{ c_1 \mathbf v_1, \dots, c_k \mathbf v_k \}$$ is a basis for $$G$$? Because if so, then $$\mathbb Z^{\oplus n} / G \cong \mathbb Z_{c_1} \oplus \dots \oplus \mathbb Z_{c_k} \oplus \mathbb Z^{n - k}$$.

[In the above example, $$n = 2$$, $$G = \langle (2, 2) \rangle$$, $$\mathcal B = \{ (1, 1), (1,0) \}$$, $$k = 1$$ and $$c_1 = 2$$.]

Yes, such an algorithm exists, and it's called the Smith normal form algorithm. Take a look at the final few pages of these notes for an explanation.

• Thank you so much!
– user752801
Jul 31, 2020 at 23:00
• I corrected the typo!
– user752801
Jul 31, 2020 at 23:00