Let G = $Z_4$ x $Z_6$ be the direct product of cyclic groups $Z_4$ and $Z_6$. Let N = <(2,3)> be a normal subgroup of G. Show that G/N $\simeq$ $Z_{12}$

What I have so far..

Given that |N| = 2, the order of $|G/N| = |G|/|N| = 24/2 = 12 = 2^{2}$*$3$

The number of abelian groups = $p^{(2)}$$p^{(1)}$ = 2

So the order 12 abelian groups are : $Z_4$ x $Z_3$ and $Z_2$ x $Z_2$ x $Z_3$

From this, I claim G/N $\simeq$ $Z_{12}$

How would I prove this from what I have so far? Thanks in advance.

  • $\begingroup$ By the Chinese Remainder Theorem, $\mathbb{Z}_{12} \cong \mathbb{Z}_{4} \times \mathbb{Z}_{3}$. Hence, one strategy would be to demonstrate an element of order $4$ in the quotient group to differentiate it from $\mathbb{Z}_{2}\times \mathbb{Z}_{2} \times \mathbb{Z}_{3}$. Alternatively, you can specify a generator to show that $G/N$ is cyclic. $\endgroup$ – Alex Wertheim Dec 14 '14 at 17:51
  • $\begingroup$ Why do you think your group is isomorphic to $\mathbb{Z}_4\times\mathbb{Z}_3$ rather than $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3$? $\endgroup$ – paw88789 Dec 14 '14 at 17:59
  • 1
    $\begingroup$ @paw88789: doesn't the element $(1, 3)$ have order $4$ in the quotient group? $\endgroup$ – Alex Wertheim Dec 14 '14 at 18:02
  • $\begingroup$ @AWertheim: Good point! I need to practice my quotient groups. $\endgroup$ – paw88789 Dec 14 '14 at 18:05
  • $\begingroup$ Sorry, I'm a bit confused. Why do I have to demonstrate that there is an element of order 4? What does this signify? Also, a silly question but why does (1,3) have order 4 and not 2? $\endgroup$ – Squires McGee Dec 14 '14 at 18:35

A straightforward consequence of Lagrange's Theorem tells us that $|G/N|=12$.

The idea now is to construct a morphism $$ \varphi\colon G\to\mathbb{Z}_{12} $$ such that $ker(\varphi)=N$. So by the first isomorphism theorem we deduce the result.

Since $G$ and $\mathbb{Z}_{12}$ are $\mathbb{Z}$-module we can extend by linearity the morphism defined by $$ (1,0)\mapsto 3,\ (0,1)\mapsto 2 $$ It is easy to check that $N=ker(\varphi)$.

  • $\begingroup$ Nice answer, +1 :) $\endgroup$ – Alex Wertheim Dec 14 '14 at 18:04
  • $\begingroup$ Yep, very elegant. Thanks everyone. $\endgroup$ – Squires McGee Dec 14 '14 at 19:22
  • $\begingroup$ Good answer. @Squires, this is a good approach to any question dealing with an isomorphism like this. The first isomorphism theory pops up all over the place. In fact, it's often called the Fundamental Homomorphism Theorem. $\endgroup$ – JeffW89 Dec 14 '14 at 21:02
  • $\begingroup$ Got it, thanks Jeff. I did go through over that theorem, didn't see it's application here until now. $\endgroup$ – Squires McGee Dec 14 '14 at 22:16

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