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$G = \Bbb Z \times \Bbb Z$ and $H = \{(a,a): a \in \Bbb Z\}$

I'm trying to find what $G/H$ isomorphic to, but I'm having a really tough time understanding this concept. Can somebody walk me through it? I figure if I have a model, it will be easier to work my way through a few problems...this seemed like the best one to choose as an example.

amWhy
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c87
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4 Answers4

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I'm assuming, based on your tags, that we are treating $G$ as an additive group. In that case, $G$ is the set of 2-tuples with entries in $\mathbb{Z}$, and the group action of addition. $H$ is a subgroup of $G$, where we are restricted to the tuples in which both entries are the same.

Hence, in $G/H$, anything in the form $(a,a)$ is equal to zero. Hence, if we take an arbitrary element $(x,y) \in G$, then, in $G/H$, we have $$(x,y) = (x,y)-(y,y) = (x-y,0)$$ because $(y,y) = 0$ in this quotient. Since we can pick $x$,$y$ so that $x-y$ is any integer, we see that this suggests that $G/H \cong \mathbb{Z}$.

Now, to formally prove this, we can consider the map from $G$ to $\mathbb{Z}$ that takes $(x,y) \to x-y$, and prove that the kernel is equal to $H$.

In general, to determine these quotients, you should try playing around with the relations given by the group you're quotienting over. Then, to make a formal argument, you should construct functions onto the space that you have determined to be the quotient, and show that $H$ is the kernel.

Dan
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  • This explanation was really helpful, thanks. Would you mind explaining the kernel bit a little more? We went over it very quickly I couldn't quite follow – c87 Jun 11 '18 at 01:04
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    Sure. If we have G,G' groups and a homomorphism $f: G \to G'$, then the kernel of $f$ are those $g \in G$ such that $f(g) = e$ where $e$ is the identity in the group $G'$. In this situation, we are working with additive groups, so we typically let 0 denote the identity. The first isomorphism theorem tells us that if we have a homomorphism $f$, then, if $H$ is the kernel of $f$, $G/H \cong G'$. Hence, if we want to prove that a quotient is isomorphic to some known group, we want to construct a function $f$ with the desired kernel, so that we can apply the isomorphism theorem. – Dan Jun 11 '18 at 01:08
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    So.. would the kernel in this case just be $x-y$? Since that would give the identity? edit I think I see. I thought there was more to be done with the kernel after what you had already explained. So then is the solution complete? That's the process I have to follow for these problems? – c87 Jun 11 '18 at 01:12
  • Well, the function $f$ we need here would send $(x,y)$ to $x-y$. The kernel would then be $H$, since every element of $H$ is in the form $(a,a)$, and $f$ maps $(a,a)$ to $a-a = 0$. But yes, this is exactly the method you can use for these problems. – Dan Jun 11 '18 at 01:27
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Consider $f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}$ defined by $f(x,y)=x-y$ show it is surjective and its kernel is $H$.

Tsemo Aristide
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One of the most natural ways to identify a quotient group $G/K$ is to find a homomorphism $\varphi: G \to H$ with $K = \ker(\varphi)$. Put simply, make your subgroup the kernel of some homomorphism. It then follows that $\text{im}(\varphi) \cong G/K$.

In your case, $G = \mathbb{Z} \times \mathbb{Z}$ and $K = \{(a,a)\,\mid\, a \in \mathbb{Z}\}$. Which function takes a pair $(a,b)$ (an element of $G$) and yields $0$ precisely when $a=b$? An obvious candidate is $\varphi((a,b))=a-b$.

Finally, just confirm that $\varphi$ is indeed a homomorphism. From where to where? What is its image?

Alon Amit
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The problem would be easy if $H= \mathbb Z e_1 = \{(a,0): a \in \mathbb Z\}$, right?

Now, $H = \mathbb Z (1,1)$. So, we try to convert the original problem into this one, by finding a basis of $G$ containing $(1,1)$. This is easy: $G = \mathbb Z \times \mathbb Z = \mathbb Z f_1 \oplus \mathbb Z f_2$ where $f_1=(1,1)$ and $f_2=(0,1)$.

Then $ H = \mathbb Z f_1 = \mathbb Z f_1 \oplus 0\mathbb Z f_2$.

Therefore, $G/H \cong \mathbb Z /\mathbb Z \times \mathbb Z/0\mathbb Z \cong \mathbb Z $.

lhf
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  • You're welcome. However, I think it would be less confusing to stick to one notation ($\times$ or $\oplus$), since they mean the same here (unless there is something I'm missing?) – Arnaud Mortier Jun 11 '18 at 01:01
  • @ArnaudMortier, $\oplus$ here means internal direct sum. I think writing $\mathbb Z f_1 \times \mathbb Z f_2$ would be confusing. – lhf Jun 11 '18 at 01:06