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Question. Let $G = \{1,2,3,4\}$. Given that $(G, \cdot)$ is a group with identity $3$ and that $o(x) = 2$ for each $x \in G \setminus \{3\}$, complete the Cayley table.

I'm trying to break apart each statement in hopes to understand how I should fill in this specific Cayley table.


"$(G, \cdot)$ is a group with identity $3$." this is pretty basic and I understand it. A group is associative, has identity ($3$) and inverses.

"$o(x) = 2$ for each $x \in G \setminus \{3\}$." This is saying that each element except ${3}$ has an order of $2$, and this is what is causing confusion. Is this saying that $x^2 = e$ (the identity), except for ${3}$? So $1\cdot1 = 3$, $2\cdot2 = 3$, $4\cdot4 = 3$? I'm guessing $3\cdot3 = 1$ (because $e\cdot e = e$)?


My second attempt thanks to everyone's help:
second attempt at cayley table

Thank you.

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    $\begingroup$ Why do you have $8$'s in your group table? $\endgroup$ – bames Feb 25 '18 at 5:34
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Yeah! This is a Klein group. this group has the same properties as $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$. The numbers $1,2,3,4$ are representations of elements.

There exist some erros in your Cayley table. In this case $1 \neq e$, $3.3 = 3$, because $3$ is identity by definition, for exemple. I suggest you make the table in order $3,1,2,4$ (start with identity, becomes more organized).

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  • $\begingroup$ thank you. I wrote the table in the order you suggested and then changed it back once I was done (It helped me see it was abelian). I didn't even think of doing that! $\endgroup$ – Charles Grealy Feb 25 '18 at 18:45
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    $\begingroup$ You're welcome! $\endgroup$ – Corrêa Feb 26 '18 at 0:52
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You understood both statements well and the conclusions you wrote are correct.

But your Cayley table is wrong. To correctly fill the table, you need to recall other properties of groups. Also, there is an 8 in your table, and 8 is not an element of the group $G$.

Hint: Let $G$ be a group and $x,y\in G$. If $xy=x$, then what can we say about $y$?

In this case, necessarily $y=e$, the identity element. Can you prove this?

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$e=3$ not $e=1$. So your last statement is incorrect. Indeed, $3\cdot 3=3\neq1$ as $e\cdot e=e$.

Similarly, in your table you say $2\cdot 1=2$ and also $2\cdot 3=2$. You know the second identity is correct, so the first cannot be for the following reason (recall that $2\cdot 2$ is the identity): $$ \begin{align*} 2\cdot 1&=2\cdot 3\\ \Leftrightarrow 2\cdot2\cdot 1&=2\cdot2\cdot 3\\ \Leftrightarrow 1&=3\ldots\\ \end{align*} $$ ...and we know that $1\neq3$. More generally, all rows and columns in every Cayley table of every group contain each element precisely once. Can you see why this is?

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    $\begingroup$ thank you for the explanation. I can see (now) why each row and column must contain each element only once, or else we would have multiple identities as you've shown. $\endgroup$ – Charles Grealy Feb 25 '18 at 18:46

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