Direct Product of Cyclic Groups and Quotient Groups

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)}$$p^{(1)} = 2 So the order 12 abelian groups are : Z_4 x Z_3 and Z_2 x Z_2 x Z_3 From this, I claim G/N \simeq Z_{12} How would I prove this from what I have so far? Thanks in advance. • By the Chinese Remainder Theorem, \mathbb{Z}_{12} \cong \mathbb{Z}_{4} \times \mathbb{Z}_{3}. Hence, one strategy would be to demonstrate an element of order 4 in the quotient group to differentiate it from \mathbb{Z}_{2}\times \mathbb{Z}_{2} \times \mathbb{Z}_{3}. Alternatively, you can specify a generator to show that G/N is cyclic. – Alex Wertheim Dec 14 '14 at 17:51 • Why do you think your group is isomorphic to \mathbb{Z}_4\times\mathbb{Z}_3 rather than \mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3? – paw88789 Dec 14 '14 at 17:59 • @paw88789: doesn't the element (1, 3) have order 4 in the quotient group? – Alex Wertheim Dec 14 '14 at 18:02 • @AWertheim: Good point! I need to practice my quotient groups. – paw88789 Dec 14 '14 at 18:05 • Sorry, I'm a bit confused. Why do I have to demonstrate that there is an element of order 4? What does this signify? Also, a silly question but why does (1,3) have order 4 and not 2? – Squires McGee Dec 14 '14 at 18:35 1 Answer A straightforward consequence of Lagrange's Theorem tells us that |G/N|=12. The idea now is to construct a morphism$$ \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)\$.

• Nice answer, +1 :) – Alex Wertheim Dec 14 '14 at 18:04
• Yep, very elegant. Thanks everyone. – Squires McGee Dec 14 '14 at 19:22
• 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. – JeffW89 Dec 14 '14 at 21:02
• Got it, thanks Jeff. I did go through over that theorem, didn't see it's application here until now. – Squires McGee Dec 14 '14 at 22:16