Suppose G has order 4, but contains no element of order 4. A) prove that no element of G has order 3.?

Question

Also, B) Explain why every non-identity element of G has order 2. C) Denote the elements of G by e, a, b, c and write out the operation table for G.

Answer 1

Work by contradiction: since we know the group has four elements, let's label them $a$, $b$, $c$ and $e$. Suppose $a$ has order 3; then $a^2\neq e$. Suppose $a^2$ is $b$ (otherwise, we could just 'swap labels' on $b$ and $c$ to make this so), and of course $a^3$ is $e$ (since we've presumed towards our contradiction that $a$ has order 3). Then what values can $ac$ have, if the group axioms are to be satisfied?

Answer 2

A relatively basic question, but one that I will answer for sake of completeness. Assuming that we have Lagrange's theorem, if $G$ has no element of order 4, then every element must have order 1 or 2. Lagrange tells us that the order of an element must divide the order of the group. Since 3 does not divide 4, our only options are 1 and 2. But what element of $G$ has order 1? Only the identity. Hence every non identity element must have order 2.

I would maybe look up the Klein-4 group.

I think that from here you can take the identity table. Note that $a^2=b^2=c^2=1$.

Answer 3

You have four elements, $\{1,a,b,c\}$. Assume one element (without loss of generality, $a$) has order $3$. Without loss of generality, assume $a^2 = b$ (we cannot have $a^2 \in \{a,1\}$ by assumption). Then $G = \{ 1,a,a^2,c\}$.

If $ac = 1$, then $c = (a^3)c = a^2(ac) = a^2$, a contradiction.

If $ac = a$, $c=1$, a contradiction.

If $ac = a^2$, $a=c$, a contradiction.

If $ac = c$, $a=1$, contradiction.

Therefore $ac \notin \{1,a,a^2,c\} = G$, so $a$ cannot have order $3$.

To write out the table, note that $G = \{1,a,b,c\}$ and that $a,b,c$ have order $2$ by what we just proved. Then $ab \notin \{1,a,b\}$ ; it is not $1$ since $a$ and $b$ are their own inverses, and it is neither $a$ or $b$ because $ab=a$ implies $b=1$, similarly for the other case. Thus $ab = c$ and similarly $ba = c$. You can then show that this trick works for the other pairs too, i.e. $ac = ca = b$ and $bc = cb = a$. I leave it up to you to write the table.

Hope that helps,