1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $N=ker(\varphi)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.