1

Let $f: \Bbb{Z} \times \Bbb{Z} \to G $ be an epimorphism, $\ker f$ be generated by $\langle (3,0),(0,5) \rangle$. What type of abelian group is $G$?

I don't even know the context of that question since I found it on the back of my abstract algebra book (without a connection to any particular topic). Any hint would be much appreciated.

EDIT (based on comment below): I know that since f is surjective then $G$ is isomorphic to $\Bbb Z\times \Bbb Z/\ker(f)$ so it is isomorphic to $\Bbb Z\times \Bbb Z/\langle(3,0),(0,5)\rangle$. I see that $3$ and $5$ are prime numbers, but I don't know what to say more.

Shaun
  • 44,997
blackkay1
  • 81
  • 1
    Is it just looking for the word finite? Or maybe the exact isomorphism class? – Dan Rust Dec 08 '21 at 16:09
  • Which book are you referring to? – Shaun Dec 08 '21 at 16:17
  • 1
    Do you know the First Isomorphism Theorem? – Shaun Dec 08 '21 at 16:34
  • I can answer this question. However, first of all, I would like you to provide context in the form of what you have tried, what similar questions you know the answer to, what you're studying, etc. Please do so in an [edit]. This site is for learning and not about answering questions with no effort on the part of the asker. – Shaun Dec 08 '21 at 16:39
  • 1
    I know that since f is surjective then G is isomorphic to $ZxZ/ ker(f)$ so it is isomorphic to $ZxZ/<(3,0),(0,5)>$. I see that 3 and 5 are prime numbers, but I don't know what to say more. – blackkay1 Dec 08 '21 at 18:02
  • @blackkay1 note that in this specific case $\langle (3,0),(0,5)\rangle=\langle 3\rangle\times\langle 5\rangle$. So can you deduce a bit more about the quotient? – freakish Dec 08 '21 at 18:15

1 Answers1

3

Since $f$ is an epimorphism, we can apply the First Isomorphism Theorem to get $(\Bbb Z\times\Bbb Z)/\ker f\cong G$.

Let $x=(1,0), y=(0,1)$. Then the presentation

$$\Bbb Z\times\Bbb Z\cong \langle x,y\mid x+y=y+x\rangle$$

is standard.

Note that $(3,0)=3x$ and $(0,5)=5y$. Then

$$\begin{align} \Bbb Z\times\Bbb Z/\ker f &\cong \langle x,y\mid 3x, 5y, x+y=y+x\rangle\\ &\cong\Bbb Z_3\times \Bbb Z_5\\ &\cong \Bbb Z_{15}, \end{align}$$

where the final isomorphism holds by the Chinese remainder theorem and the fact that $\gcd(3,5)=1$.

Shaun
  • 44,997
  • 1
    Why is that presentation a standard? What if I had different generators that would not be only prime numbers and zeros? Let's say that those would be $\langle (3,1),(8,2) \rangle$ and not $\langle (3,0),(0,5) \rangle$. – blackkay1 Dec 08 '21 at 18:25
  • It is standard, @blackkay1, because $\Bbb Z\cong\langle a\mid \rangle$ as a free group of rank one, and the direct product of two presentations is as given here. And that's a separate question; if you're interested, I suggest you make a new post on it. – Shaun Dec 08 '21 at 18:29
  • So you say that doing the same thing for different generators would not be that totally similar to that example? – blackkay1 Dec 08 '21 at 18:38