Classifying Quotient Groups

Question

We are going over classifying quotient groups in my intro group theory class, and this is the first thing that has given me genuine trouble in the class (and boy, has it given me trouble). The case of finite quotient groups makes sense to me, Lagrange's theorem makes everything work out nicely, but as soon as we work with infinite groups, I have no idea what is happening. I have scoured this site, and found one post (here) that has helped me a lot, but I don't know how to generalize it further, and the textbook's (A First Course in Abstract Algebra) explanations of similar examples just don't make sense to me.

When we look at the case of $\mathbb{Z}\times \mathbb{Z} \times \mathbb{Z}/\langle(1,1,1)\rangle,$ the method explained in the linked post makes perfect sense. Then in a case such as $\mathbb{Z}\times\mathbb{Z}/\langle(2,2)\rangle$, I think it is fair to say that every coset will contain either an element $(m,0)$ or $(m,1)$, and use this to say it is isomorphic to $\mathbb{Z}\times\mathbb{Z}_2$. But what about $\mathbb{Z}\times\mathbb{Z}/\langle(2,6)\rangle$? When the elements of the ordered pair of the generator are not equal, I am not so sure what to do. My intuition is to say that "each element in the given quotient group will contain an element of the form $(p,m+6k)$ for some $m,k \in \mathbb{Z}$, and some $p:0\leq p<2$. Additionally, it must contain an element of the form $(n+2s,q)$, for some $n,s \in \mathbb{Z}$, and some $q:0,\leq q < 5.$ Thus, the given group is isomorphic to $\mathbb{Z}_6 \times \mathbb{Z}_2$." But I do not trust this reasoning at all, something feels very shaky, and it doesn't help that I am having a bit of a difficult time visualizing these groups to begin with. I have heard of Smith normal form on this board, but I don't think that would be accepted on my imminent exam. Any help would be greatly appreciated.

Answer 1

The quotient $\mathbf Z^2/\langle(2,6)\rangle$ is definitely not $\mathbf Z_6 \times \mathbf Z_2$ and is not even finite: when $H$ is a subgroup a $\mathbf Z^n$ with free rank $m$, the quotient group $\mathbf Z^n/H$ has free rank $n - m$. Thus $\mathbf Z^2/\langle(2,6)\rangle$ has free rank $2-1 = 1$ and $\mathbf Z^n/H$ is finite if and only if $H$ has free rank $n$.

A vector $v \in \mathbf Z^n$ is part of a basis of $\mathbf Z^n$ if and only if $v$ is primitive: its coefficients have gcd $1$. And $n$ vectors $v_1, \ldots, v_n$ in $\mathbf Z^n$ are a basis if and only if the $n \times n$ matrix with the $v_i$'s as columns has determinant $\pm 1$.

Thus when trying to determine what $\mathbf Z^2/\langle(2,6)\rangle$ is in a concrete way, write $(2,6)$ as $2(1,3)$ and realize the primitive vector $(1,3)$ as part of a basis of $\mathbf Z^2$, say $\{(0,1),(1,3)\}$ (but not $\{(1,0),(1,3)\}$). Thus $\mathbf Z^2 = \mathbf Z(0,1) \oplus \mathbf Z(1,3)$, so $$ \mathbf Z^2/\langle(2,6)\rangle = \mathbf Z^2 / \mathbf Z2(1,3) = (\mathbf Z(0,1) \oplus \mathbf Z(1,3))/ \mathbf Z2(1,3) \cong \mathbf Z \oplus \mathbf Z_2. $$

I'd determine the other examples you listed in a similar way. For $\mathbf Z^2/\langle(2,2)\rangle$, write $(2,2)$ as $2(1,1)$ and realize the primitive vector $(1,1)$ as part of the basis $\{(1,0),(1,1)\}$ of $\mathbf Z^2$. Thus $$ \mathbf Z^2/\langle(2,2)\rangle = \mathbf Z^2 / \mathbf Z2(1,1) = (\mathbf Z(1,0) \oplus \mathbf Z(1,1))/ \mathbf Z2(1,1) \cong \mathbf Z \oplus \mathbf Z_2. $$ And since $(1,1,1)$ is part of the basis $\{(1,0,0),(0,1,0),(1,1,1)\}$ of $\mathbf Z^3$, $$ \mathbf Z^3/\langle(1,1,1)\rangle = \mathbf Z^3 / \mathbf Z(1,1,1) = (\mathbf Z(1,0,0) \oplus \mathbf Z(0,1,0) \oplus \mathbf Z(1,1,1))/ \mathbf Z(1,1,1) \cong \mathbf Z^2. $$

Remark. When $(a,b)$ is an arbitrary nonzero vector in $\mathbf Z^2$, we can use the same ideas to determine the structure of $ \mathbf Z^2/\langle(a,b)\rangle$. Let $d = \gcd(a,b)$, $a = da'$, and $b = db'$. Then $(a,b) = d(a',b')$ and $(a',b')$ is a primitive vector, so $(a',b')$ is part of a basis of $\mathbf Z^2$, say $\{(k,\ell),(a',b')\}$. Thus $$ \mathbf Z^2/\langle(a,b)\rangle = \mathbf Z^2 / \mathbf Zd(a',b') = (\mathbf Z(k,\ell) \oplus \mathbf Z(a',b'))/ \mathbf Zd(a',b') \cong \mathbf Z \oplus \mathbf Z_d. $$ In a similar way, when $v$ is an arbitrary nonzero vector in $\mathbf Z^n$ and $d$ is the gcd of the coefficients of $v$, $\mathbf Z^n/\langle v\rangle \cong \mathbf Z^{n-1}\oplus \mathbf Z_d$.

Answer 2

Given a quotient group $G / N$, one approach is to find $H$ and a surjective homomorphism $\phi : G \to H$ such that $\ker \phi = N$. Then the first isomorphism theorem implies $G/N \simeq H$.

• For $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle(1, 1,1)\rangle$, we want a homomorphism $\phi$ on $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$ such that $\ker \phi = \langle(1, 1, 1)\rangle$. One option is $\phi : \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ given by $\phi((x,y,z)) = (y-x, z-x)$. (See how $\phi$ encodes how "all components are the same" in $\langle(1, 1, 1)\rangle$.) You can check that $\ker \phi = \langle(1, 1, 1)\rangle$ and that $\phi$ is surjective.
• For $\mathbb{Z} \times \mathbb{Z} / \langle(2, 2)\rangle$, we want a homomorphism $\phi$ on $\mathbb{Z} \times \mathbb{Z}$ such that $\ker \phi = \langle(2,2)\rangle$.
  • Using $\phi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ given by $\phi((x, y)) = y - x$ yields $\ker \phi = \langle(1, 1)\rangle$ which is too large: we don't want the elements $(u, u)$ where $u$ is odd.
  • To get rid of these elements, we can instead use $\phi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}_2$ given by $\phi((x, y)) = (y-x, x \mod 2)$. You can check that $\ker \phi$ is exactly $\langle(2, 2)\rangle$, and that $\phi$ is surjective.
• For $\mathbb{Z} \times \mathbb{Z} / \langle(2, 6)\rangle$, we want a homomorphism $\phi$ on $\mathbb{Z} \times \mathbb{Z}$ such that $\ker \phi = \langle(2, 6)\rangle$.
  • Similar to before, $\phi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ given by $\phi((x, y)) = y - 3x$ almost works, but the kernel $\langle(1, 3)\rangle$ contains too many elements.
  • We can fix this in a similar way by using $\phi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}_2$ given by $\phi((x, y)) = (y-3x, x \mod 2)$. You can check that $\ker \phi$ is exactly $\langle(2, 6)\rangle$, and that $\phi$ is surjective.