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In Fraleigh's example he states: enter image description here

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.

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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.

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  • $\begingroup$ So is the possible of order of elements in Z2: 1 and 2 and the possible order of elements in Z3: 1,2,3? $\endgroup$ – rain Mar 20 '18 at 9:46
  • $\begingroup$ I see, thanks, makes perfect sense. Thanks. $\endgroup$ – rain Mar 20 '18 at 9:50
  • $\begingroup$ @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. $\endgroup$ – Kaj Hansen Mar 20 '18 at 11:19
  • $\begingroup$ Thanks, makes sense. $\endgroup$ – rain Mar 21 '18 at 1:03
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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 $2$.

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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.

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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 $4$.

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