# What is $\mathbb{Z}_2 \times \mathbb{Z}_4$ isomorphic to - Fraleigh p. 112 Exercises 11.32e

(e). p. 4 of PDF - $\mathbb{Z}_2 \oplus \mathbb{Z}_4 \not\simeq \mathbb{Z}_8$. Another solution

(1.) Why is $\mathbb{Z}_2 \oplus \mathbb{Z}_4$ not cyclic? Is it because of $\gcd(2, 4) = 2 \neq 1 \iff$ p. 106 Corollary 11.6?

Reference: p. 11 of the same PDF. robjohn nicely the version with no colors under. original image.

Alternative Approach to showing $f$ is onto:
Define $f:\mathbb{Z}_6\to\mathbb{Z}_2\oplus\mathbb{Z}_3$ by $f(m)=(m\mod2,m\mod3)$.

$2$ and $3$ relatively prime $\implies\exists s,t$ such that $1=2s+3t$. In fact, $1=2(-1)+3(1)$.
Let $(a,b)\in\mathbb{Z}_2\oplus\mathbb{Z}_3$. Need to find a $g\in\mathbb{Z}_6$ such that $f(g)=(a,b)$.

Define $\color{magenta}{g=3ta+2sb}$. That is, define $g=3(1)a-2(-1)b=3a-2b$. NTS $f(g)=(a,b)$.

(2.) Where does this $\color{magenta}{g=3ta+2sb}$ spring from magically? Can someone please unfold it?

• Please use \mathbb correctly (i.e., only on the specific capital letters you want it to apply to, not an entire expression.) – Zev Chonoles Feb 21 '14 at 9:08

Corollary 11.6 you describe exactly tells you that $\mathbb Z_2\oplus \mathbb Z_4$ is not cyclic, so your reasoning is correct. Another way of proving that is simple.
Take any element $a\oplus b \in\mathbb Z_2\oplus \mathbb Z_4$ Now it's easy to show that $$(a\oplus b)+(a\oplus b)+(a\oplus b)+(a\oplus b) = 0,$$ meaning all elements of the group have order at most $4$.
Since $1\in\mathbb Z_8$ has order $8$, the groups are not isomorphic.
For the second question, the simplest way to answer is that picking that $g$ simply works. More completely thought, the idea is not that hard.
You want to find such a $g$ that $$g\equiv a \mod 2$$ and $$g\equiv b \mod 3.$$
Setting $g=3a+2b$ makes sure both equations hold.
• Thanks. But the answer has s and t for its $\color{magenta}{g=3ta+2sb}$. Can you please unfold why yours doesn't? – Group Theory Feb 22 '14 at 12:46