0

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.

Stahl
  • 23,855
Koro
  • 11,766
  • 1
    You cannot start by assuming that the order of $G$ is $m,$ as $G$ may be (and in fact is) infinite. – Stahl Jul 09 '20 at 20:49
  • @Stahl: Why? I thought about disproving my assumption using contradiction so I assumed that. Anything wrong with it? – Koro Jul 09 '20 at 20:50
  • 2
    Ah, I'm sorry; I missed that you use this for contradiction! Then I'd make a stylistic comment -- when you start by assuming something you want to contradict, it helps the reader to write something like "assume for the sake of contradiction that..." so the reader does not think you're making an unproven assumption. – Stahl Jul 09 '20 at 20:51
  • @Stahl: Will edit it right now – Koro Jul 09 '20 at 20:53
  • 2
    The operation in $\mathbb{Z}+\mathbb{Z}$ is addition, so cosets are not $(m,n)<(2,2)>$ but $(m,n)+<(2,2)>$ – markvs Jul 09 '20 at 21:41
  • @AnuragA: No it doesn't. My question is verification while the link is for finding isomorphism from G. Both are completely different things. – Koro Jul 09 '20 at 22:15
  • 1
    These are related. the old question gives the answer to the new one. I have deleted my answer and voted to close the questiom. – markvs Jul 09 '20 at 22:32
  • 1
    @JCAA Related, but an answer to one does not give an answer to the other, and in particular, the question linked does not answer the question asked here. The question here asks if the provided argument is a proof, the other asks for a proof of an isomorphism. A specific proof does not answer the question of whether the provided argument holds, and answering if the argument is correct does not necessarily provide a proof of the isomorphism (although it may). Of course, one may object to "proof-verification" type questions, but that's a meta-issue. – Stahl Jul 09 '20 at 22:36

2 Answers2

2

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.

Stahl
  • 23,855
  • Wow. Thanks a lot. I understood. – Koro Jul 09 '20 at 21:38
  • I would like to clarify my argument about proving G is not cyclic. I think it was unclear in my post. I assumed $H=\langle (2,2) \rangle$, which is a subgroup of $Z\oplus Z$. So $(a,b)\langle (2,2) \rangle$ and $(1,1)\langle (2,2) \rangle$ are two cosets of H and hence must be of same order. – Koro Jul 09 '20 at 21:42
  • I see a little more clearly what you mean. However, you are conflating cosets in $\Bbb{Z}\oplus\Bbb{Z}$ with cosets in $G.$ As cosets in $\Bbb{Z}\oplus\Bbb{Z}$, these do have the same number of elements -- they're both infinite, so there's no contradiction – Stahl Jul 09 '20 at 21:45
  • Order of identity is one and not zero. I'm thinking about the rest part of your comment. – Koro Jul 09 '20 at 22:08
  • @Koro You're absolutely right. I've reposted the comment to correctly state what I mean, as it can no longer be edited. – Stahl Jul 09 '20 at 22:10
  • 1
    Moreover, it seems you're conflating the cardinality of cosets in $\Bbb{Z}\oplus\Bbb{Z}$ with the order of the cosets in the quotient. It will generally not be the case that given a group $G$ and a normal subgroup $H$, that two cosets $aH$ and $bH$ of $H$ will have the same order in $G/H$. For example, the trivial coset $H$ itself always has order 1, because it's the identity of $G/H$. But if the quotient group is nontrivial, then some nontrivial coset has order $>1$. And I hope you agree that not all quotient groups are the trivial group! – Stahl Jul 09 '20 at 22:12
  • 1
    I got it. So basically order of elements in factor groups G/H are not the same. And I confused these with cosets of H in G. This is really confusing though as G/H is set of all cosets of H in G... But this is probably because identities of G and G/H are different. Moreover, G/H is not a subset of G. – Koro Jul 09 '20 at 22:18
  • 1
    @Koro Right: the cardinality of cosets of $H$ in $G$ tells you nothing about the order of the coset as an element of $G/H.$ All cosets have the same cardinality, but orders of elements in $G/H$ can vary wildly -- take any group $G,$ and remember that $G\cong G/{e}.$ – Stahl Jul 09 '20 at 22:21
0

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.

Dylan C. Beck
  • 4,933