Find the Fallacy in the Factor Group Logic

Question

The factor group for $(\Bbb{Z} \times \Bbb{Z})/\langle(2, 0)\rangle$ is clearly $\Bbb{Z} \times \Bbb{Z}_2$ because all of the elements are in the form of $a(1, 0)+b(0, 1)+\langle(2, 0)\rangle$ for $a \in \Bbb{Z}_2$ and $b \in \Bbb{Z}$. However, I also have the following argument that it is $\Bbb{Z} \times \Bbb{Z}_2 \times \Bbb{Z}_2$ and I do not know where the mistake is.

$$\tag1 \frac{\langle(2, 0), (0, 2)\rangle}{\langle(2, 0)\rangle} \cong \Bbb{Z}$$ because all of the elements are in the form of $z(0, 2)+\langle(2, 0)\rangle$ for $z \in \Bbb{Z}$.

$$\tag2 \frac{\Bbb{Z} \times \Bbb{Z}}{\langle(2, 0), (0, 2)\rangle}=\Bbb{Z}_2 \times \Bbb{Z}_2$$

because these are the following elements:

• $\langle(2, 0), (0, 2)\rangle$
• $(1, 0)+\langle(2, 0), (0, 2)\rangle$
• $(0, 1)+\langle(2, 0), (0, 2)\rangle$
• $(1, 1)+\langle(2, 0), (0, 2)\rangle$

$$\tag3 \frac{\Bbb{Z} \times \Bbb{Z}}{\langle(2, 0), (0, 2)\rangle} \times \frac{\langle(2, 0), (0, 2)\rangle}{\langle(2, 0)\rangle}=\Bbb{Z} \times \Bbb{Z}_2 \times \Bbb{Z}_2$$

However, I am pretty sure this is wrong because it does not make sense. Can someone please explain my mistake? Thanks!

Answer 1

You used the following formula:

$$\frac G N*\frac N H \cong \frac G H$$

However, this was your flaw because this is not necessarily true all of the time, as shown by this counter example. In fact, given that $N$ is a normal subgroup of $G$ and $H$ is a normal subgroup of $N$, it is not even guaranteed $H$ is a normal subgroup of $G$, so given $G/N$ and $N/H$, one can not even deduce $G/H$ exists. However, you can use the Third Isomorphism Theorem, which states the following (given that $N, K$ are normal subgroups of $G$ and that $N \subseteq K \subseteq G$):

$$\frac{\frac G N}{\frac K N} \cong \frac G K$$

There is also the Second Isomorphism Theorem, which states the following (given $H, N$ are subgroups of G and $N$ is normal):

$$\frac{HN}{N} \cong \frac{H}{H \cap N}$$

You can use one of these valid rules to help you prove factor groups, but you can not use the invalid rule of just cancelling out groups because it does not always work, as shown by this counterexample.

Also, looks can be deceiving. You have not proven that $(\Bbb{Z} \times \Bbb{Z})/\langle(2, 0)\rangle$ simply from your formula. You can also construct a formula such as $a(1, 0)+b(0, 1)+c(0, 2)+\langle(2, 0)\rangle$ for $a, b \in \Bbb{Z}_2$ and $c \in \Bbb{Z}$. However, it is not actually isomorphic to $\Bbb{Z} \times \Bbb{Z}_2 \times \Bbb{Z}_2$:

$$((0, 1)+2(0, 2)+\langle(2, 0)\rangle)+((0, 1)+3(0, 2)+\langle(2, 0)\rangle)=2(0, 1)+5(0, 2)+\langle(2, 0)\rangle=6(0, 2)+\langle(2, 0)\rangle$$

Here, $(a_1, b_1, c_1)=(0, 1, 2)$ and $(a_2, b_2, c_2)=(0, 1, 3)$, yet $(a_1, b_1, c_1)+(a_2, b_2, c_2)=(0, 0, 6)$ instead of $(0, 0, 5)$ like in $\Bbb{Z}_2 \times \Bbb{Z}_2 \times \Bbb{Z}$, so even though it looks isomorphic because of my formula, it is actually not. Your formula is actually isomorphic to $\Bbb{Z}_2 \times \Bbb{Z}$, but just make sure that you prove the isomorphism somehow like @Joanpemo did using the First Isomorphism Theorem.

Answer 2

Here's an idea that, I believe, will solve the problem and make it swiftly. Define

$$F:\Bbb Z\times \Bbb Z\to\Bbb Z_2\times\Bbb Z\;,\;\;f(a,b)=\left(a\pmod2,b\right)$$

Check the above is a surjective group homomorphism and its kernel is

$$(a,b)\in\ker F\iff a\in2\Bbb Z\;\;\wedge\;\;b=0\iff (a,b)\in\langle\,(2,0)\,\rangle$$

Answer 3

You are trying to use a "cancellation law": $$\frac{A}{B}\times \frac{B}{C} = \frac{A}{C}.$$ Over, say, complex numbers (with non-zero denominators) this is fine, but when this is over groups it's nonsense. There's no way to cancel like this over a direct product of groups (even if we replace the equality with an isomorphism). In your case you're trying to do this with $A=\mathbb Z\times\mathbb Z$, $B=\langle (2,0),(0,2)\rangle$, and $C=\langle 2,0\rangle$.