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Let $G = \mathbb{Z}_{12} \times \mathbb{Z}_{12}$, and let $a$ be a generator of $\mathbb{Z}_{12}$.

Consider the subgroup $H$ generated by $(a^4, a^6)$. I need to write $G/H$ as a product of cyclic groups each of which has order equal to a power of some prime.

My work

Here $|H| =6$, so by Lagrange's theorem we have $|G/H|=\dfrac{144}{6}=24=8\times 3 = 2^3 \times 3$.

Can I say that $G/H \cong \mathbb{Z}_8 \times \mathbb{Z}_3$?

Shaun
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Adam_math
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    Would you consider $\mathbb Z_4 \times \mathbb Z_2 \times \mathbb Z_3$ as a possibility? What about $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$? – Lee Mosher Jul 15 '23 at 13:31
  • @LeeMosher, should I calculate the elements of $G/H$ to determine which one is the case? – Adam_math Jul 15 '23 at 14:38
  • What you are asked to do is to construct an isomorphism from $G/H$ to a group as asked. – Lee Mosher Jul 15 '23 at 14:45
  • So, not just to *identify* which one (by whatever means, perhaps your idea of a calculation would be fine), but then, once you have identified the correct group, to *write $G/H$ as* that group, which formally means to construct an isomorphism. – Lee Mosher Jul 15 '23 at 14:57
  • @LeeMosher, is there elegent way to construct this isomorphism and prove it? Any hint or theorem to apply? – Adam_math Jul 15 '23 at 15:21
  • Note that the square of the generator of $H$ is $(a^8,1)$. Hence it contains its square, which is $(a^4,1)$. Thus, $H$ is also generated by $(a^4,1)$ and $(1,a^6)$. That makes computing $G/H$ much simpler. Your argument is invalid /incomplete because the group you list is not the only abelian group of that order. – Arturo Magidin Jul 15 '23 at 15:45
  • @ArturoMagidin, So, the required map $\phi: G \rightarrow \mathbb{Z}_3 \times \mathbb{Z}_8$ defined by $\phi(a,a)=(a^3,a^2)$. Then I only need to show that $\phi$ is surjective homo then by first isomorphism theorem, we have $G/H$ and $\mathbb{Z}_3 \times \mathbb{Z}_8$ are isomorphic, right? – Adam_math Jul 15 '23 at 16:52
  • $(a,a)$ does not generate $G$. You have not defined a morphism yet. – Arturo Magidin Jul 15 '23 at 18:10

1 Answers1

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By this standard result, a presentation for $G$ is

$$\langle x,y\mid x^{12}, y^{12}, xy=yx\rangle.$$

The element that generates $H$ corresponds to $x^4y^6$, so that, by definition of a presentation, we have

$$G/H\cong\langle x,y\mid x^{12}, y^{12}, xy=yx, x^4y^6\rangle.$$

Now we can perform Tietze transformations:

$$\begin{align} G/H&\cong\langle x,y\mid x^{12}, y^{12}, xy=yx, x^4=y^{-6}=y^6\rangle\\ &\cong \langle x,y\mid x^{12}, (y^6)^2, xy=yx, x^4=y^6\rangle\\ &\cong\langle x,y\mid x^{12}, (x^4)^2, xy=yx, x^4=y^6\rangle\\ &\cong\langle x,y\mid x^{\gcd(12,8)}=x^4=e, y^6=x^4, xy=yx\rangle\\ &\cong\langle x,y\mid x^4, y^6, xy=yx\rangle\\ &\cong\Bbb Z_4\times\Bbb Z_6\\ &\cong \Bbb Z_{12}\times\Bbb Z_2, \end{align}$$

where the final isomorphism is because $\Bbb Z_{12}\cong\Bbb Z_4\times\Bbb Z_3$ while $\Bbb Z_6\cong\Bbb Z_3\times \Bbb Z_2$.

Shaun
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    Thank youf or your answer, but I couldn't understand what you mean by the last condition in the presentation of $G/H$, in particular, $x^4 y^6$. – Adam_math Jul 16 '23 at 01:07
  • Think of $x=(a,0)$ and $y=(0,b)$, @Adam_math. Then $x^4y^6=(a^4,b^6)$. For any group $L\cong\langle X\mid R\rangle$ for sets of generators $X$ and relations $R$, if $M=\langle\langle S\rangle\rangle$ is the normal subgroup generated by a set of relations $S$, then $$L/M\cong\langle X\mid R\cup S\rangle.$$ – Shaun Jul 16 '23 at 11:40
  • This might help, @Adam_math. – Shaun Jul 16 '23 at 11:47
  • The last condition, @Adam_math, means $$x^4y^6=e.$$ – Shaun Jul 16 '23 at 11:52
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    thank you very much for the clarification. How do you know that the subgroup $H$ is normal? is that because it is cyclic? – Adam_math Jul 16 '23 at 15:02
  • can we write the final isomorphism in terms of cyclic groups whose orders are powers of some primes? – Adam_math Jul 16 '23 at 15:03
  • Every subgroup of an abelian group is normal, @Adam_math. And yes: $$G/H\cong \Bbb Z_4\times \Bbb Z_3\times\Bbb Z_2.$$ – Shaun Jul 16 '23 at 15:14