# Order of element in a group

In Fraleigh's example he states:

I get pretty much everything that is happening, except I am confused by one fact. How does he know he know that $\mathbb Z_4 \times \mathbb Z_3$ has an element of order $4$ while $\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_3$ does not? I know the maximum possible order of an element in $\mathbb Z_4 \times \mathbb Z_3$ is $\text{lcm}(4,3)=12$ while the maximum possible order of element in $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$ is $\text{lcm}(2,2,3)=6$. How do I specifically know just by looking at $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$, that it does not have an element of order $4$. Is there a theorem which states the different possible orders of elements of a group such as $\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_3$? Or would I just have to write down all the possible elements of each group and calculate the order by hand of each element? Thank you.

You're quite close in recognizing that least common multiples play a central role. The key fact is that if we have a product group $\displaystyle G = \displaystyle \prod_{k=1}^n G_k$, then the order of an element $(x_1, ..., x_n) \in G$ is precisely $\text{lcm}(|x_1|, ..., |x_n|)$ where $|x_k|$ denotes the order of $x_k$ in its respective group. So if we want to determine the possible orders of elements in $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3$, we just need to consider the possible orders of elements in $\mathbb{Z}_2$ and in $\mathbb{Z}_3$ and see if the least common multiple of any triplet of possibilities can be $4$.

If we were to take a more naive approach by trying to find the order of each element by hand, it would be perhaps easier to recognize the isomorphism$^\dagger$ $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_2 \times \mathbb{Z}_6$ before one begins (fewer coordinates to deal with). However, as one attempts to find the order of each element in the latter product manually, it quickly becomes apparent, even to a previously-uninformed student, that the rule discussed above holds!

$^\dagger$This comes from this fact.

• So is the possible of order of elements in Z2: 1 and 2 and the possible order of elements in Z3: 1,2,3? – rain Mar 20 '18 at 9:46
• I see, thanks, makes perfect sense. Thanks. – rain Mar 20 '18 at 9:50
• @rain, amidst my tiredness, I made an error. You are correct about the order of elements in Z2, but the order of an element in Z3 is going to be either $1$ (in the case of the identity) or $3$ since $1+1 = 2 \quad \rightarrow \quad 2+1 = 0$ and $2+2=1 \quad \rightarrow \quad 1+2 =0$. This doesn't change the fundamental argument though. – Kaj Hansen Mar 20 '18 at 11:19
• Thanks, makes sense. – rain Mar 21 '18 at 1:03

That's quite simple: a triple $(x,y,z)$ satisfies $4(x,y,z)=(4x,4y,4z)=0$ if and only if $$\begin{cases}4x=0\\4y=0\\4z=0\end{cases}.$$ However $4$ is a unit mod.3, so $4z=0$ implies $z=0$, and you element is $(x,y,0)$. But $x$ and $y$ have order $1$ or $2$, so $(x,y,0)$ has order $1$ or $2$.

Proof by direct calculation is quite simple here. Let $(a,b,c) \in Z_2 \times Z_2 \times Z_3$. Then

$4(a,b,c) = (0,0,0)$

$\Rightarrow (4a,4b,4c)=(0,0,0)$

$\Rightarrow 4c = 0 \mod 3$

$\Rightarrow c=0 \mod 3$

but any $(a,b,0) \in Z_2 \times Z_2 \times Z_3$ has an order of 1 or 2, not 4. So no element of $Z_2 \times Z_2 \times Z_3$ has an order of 4.

Not only the maximum possible order of an element in $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$ is lcm $(2,2,3)=6$, but we actually have $6a=0$ for all $a \in \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$. Therefore, the order of every element divides $6$ and so cannot be $4$.