Verification: Determining the order of factor group $G=(\mathbb Z \oplus \mathbb Z)/\langle(2,2)\rangle$. Is the group cyclic?

Question

I proceeded as follows:

Let's assume on the contrary that order of $G$ is finite i.e., $|G|=m$ for some $m \in \mathbb N $. Hence, for any $(x,y)\in \mathbb Z \oplus \mathbb Z$, we have that: $m((x,y)+\langle (2,2) \rangle)=e_G$, where $e_G=\langle (2,2) \rangle$ is the identity of $G$.

Hence, \begin{align*} m((x,y) + \langle (2,2) \rangle)&= m(x,y) + \langle (2,2) \rangle\\ &=(mx,my) +\langle (2,2) \rangle\\ &=\langle (2,2) \rangle \end{align*} Since $(1,0)\in \mathbb Z \oplus \mathbb Z$, we must have \begin{align*} (m,0) +\langle (2,2) \rangle&=\langle (2,2) \rangle\\ \implies (m,0)&\in \langle (2,2) \rangle\\ \implies (m,0)&=(2r,2r) \end{align*} for some $r\in \mathbb Z$. Hence, $r=0$ whence $m=0$ which is a contradiction. Hence, $|G|$ is infinite.

Now for the cyclic part, let $G$ be, on the contrary, cyclic. Hence, let $G=\langle (a,b)+ \langle (2,2) \rangle \rangle$ for some $(a,b)\in \mathbb Z\oplus \mathbb Z$. Thus there must exist an element of infinite order in $G$. We have: \begin{align*} 2((1,1) +\langle (2,2) \rangle )&=(1,1)+(1,1)+ \langle (2,2) \rangle\\ &=(2,2)+\langle (2,2) \rangle\\ &=\langle (2,2) \rangle\\ &=e_G\\ \implies |(1,1) + \langle (2,2) \rangle|&=2. \end{align*} But all cosets have same no. of elements. That is, in particular, we must have, $$|(1,1) +\langle (2,2) \rangle|=2=|(a,b) +\langle (2,2) \rangle|,$$ which is a contradiction as the element on right hand side must have an infinite number of elements. Therefore, $G$ is not cyclic.

Is my proof above correct? Please let me know. Thanks.

Answer 1

First, your argument that $G$ is infinite is correct. Nice work.

However, your argument about cosets is incorrect, or at least unclear. In particular, the subgroups generated by $(1,1) + \langle(2,2)\rangle$ and $(a,b) + \langle (2,2)\rangle$ need not be cosets (of each other) -- and in fact, they aren't. Two different subgroups are never cosets of each other!

More generally, given any group $G$ and any subgroup $H\subseteq G,$ the cardinalities of any two cosets of $H$ are the same; that is, for any $a,b\in G,$ we have $\left|aH\right| = \left|bH\right|$. (This holds even when $H$ is infinite.) However, it is absolutely not the case that we will have $\left|H'\right| = \left| H\right|$ for $H',H\subseteq G$ two different subgroups -- and it appears that this is what you are claiming.

However, you could argue something like the following, which is perhaps the intuition you had: suppose for the sake of contradiction that $G$ is cyclic, with generator $(a,b) + \langle(2,2)\rangle$. Then we must have that $m(a,b) + \langle(2,2)\rangle = (1,1) + \langle(2,2)\rangle$ for some $m\in\Bbb{Z}.$ We know that $G$ is infinite, so $(a,b) + \langle(2,2)\rangle$ has infinite order. But on the other hand, \begin{align*} 2m(a,b) + \langle(2,2)\rangle &= 2(1,1) + \langle(2,2)\rangle\\ &= (2,2) + \langle(2,2)\rangle\\ &= \langle(2,2)\rangle \end{align*} which is the identity of $G.$ This implies that the order of $(a,b) + \langle(2,2)\rangle$ divides $2m,$ which is a contradiction.

Let me also make a stylistic remark. You're mixing the additive notation for a group with the multiplicative notation, which should not be done. In particular, when a group law is written additively, then an element $g$ repeated $m$ times is denoted $mg,$ not $g^m.$ In abelian groups, we often write the group law additively, and in this case we certainly would, as the group law on $\Bbb{Z}\oplus\Bbb{Z}$ is addition in both coordinates. Similarly, we denote an element in the quotient group $\Bbb{Z}\oplus\Bbb{Z}/\langle(2,2)\rangle$ by $(a,b) + \langle(2,2)\rangle,$ not $(a,b)\langle(2,2)\rangle,$ and we would denote a coset of a subgroup $H$ of $\Bbb{Z}\oplus\Bbb{Z}$ by $(a,b) + H,$ not $(a,b)H.$

Finally, you had a question about a comment of mine saying that $\Bbb{Z}$ is the only cyclic group of infinite order up to isomorphism. What I mean by this is that if $G$ is a cyclic group and $G$ is infinite, then $G\cong\Bbb{Z}.$ To prove this, let $g\in G$ be a generator. Then the map $g\mapsto 1$ is an isomorphism between $G$ and $\Bbb{Z}$ (exercise: prove this!). This implies that if you have an infinite cyclic group $G,$ you may assume that it is $\Bbb{Z}.$ The relevance of this here is that $\Bbb{Z}$ has no element of order $2,$ so your group cannot be cyclic, as an isomorphism between groups preserves the order of elements.

Answer 2

Consider the linear transformation $\varphi : \mathbb Z \oplus \mathbb Z \to \mathbb Z \oplus \mathbb Z $ defined by $\varphi(x, y)^T = A(x, y)^T,$ where $A$ is the $1 \times 2$ matrix $A = (2 \, \, \, \, 2)$ over $\mathbb Z.$ Considering that $\mathbb Z$ is a principal ideal domain, there exists an invertible $1 \times 1$ matrix $P$ and an invertible $2 \times 2$ matrix $Q$ such that $A = PDQ,$ where $D$ is a $1 \times 2$ matrix whose diagonal entry is nonzero and whose non-diagonal entry is zero. We refer to the matrix $D$ as the Smith Normal Form of $A,$ and we obtain the matrix $D$ by performing elementary row and column operations on $A$ (corresponding to the matrices $P$ and $Q,$ respectively).

Our job in this case is not so difficult: we have that $D = (2 \, \, \, \, 0)$ by performing the elementary column operation $C_2 - C_1 \to C_2$ in $A,$ i.e., the column operation to subtract column 1 from column 2 and replace column 2 by this new column. Consequently, we have that $$\frac{\mathbb Z \oplus \mathbb Z}{\langle (2, 2) \rangle} = \operatorname{coker} \varphi \cong \frac{\mathbb Z \oplus \mathbb Z}{\langle (2, 0) \rangle} \cong \frac{\mathbb Z}{2 \mathbb Z} \oplus \mathbb Z,$$ where the last isomorphism is given by the First Isomorphism Theorem applied to the group homomorphism $\psi : \mathbb Z \oplus \mathbb Z \to \mathbb Z / 2 \mathbb Z \oplus \mathbb Z$ defined by $\psi(m, n) = (m \operatorname{ (mod } 2), n).$

Consequently, the factor group in question is infinite, and it is not cyclic because it is not isomorphic to $\mathbb Z,$ the unique (up to isomorphism) infinite cyclic group. In order to see this fact, consider an infinite cyclic group $G = \langle g \rangle,$ and define a group homomorphism $\gamma : G \to \mathbb Z$ by declaring that $\gamma(g^k) = k.$ Observe that this is surjective since for each integer $n$ in $\mathbb Z,$ its preimage in $G$ is simply $g^n.$ Further, the kernel of this map is the identity $e_G,$ so it is injective.