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I need to prove there is no group that contain only one element of order 3.

im don't know where to start because I don't really know anything about the group.

will appreciate your help

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3 Answers 3

up vote 6 down vote accepted

Hint: Suppose $a\in G$ has order $3$. What is the order of $a^2$?

EDIT: Since $a$ has order $3$, we know that $a^3=e$, the identity element. This means we can calculate $$a^2=a^2,\quad a^2\cdot a^2=a^4=a^3\cdot a=e\cdot a=a,\quad a^6=(a^3)^2=e^2=e.$$ Hence, the order of $a^2$ is $3$.

The only problem might be if $a^2=a$, but if that is the case, then we can multiply both sides by $a^{-1}$ and see that $a=e$, which is a contradiction since the order of $a$ is $3$, not $1$.

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how can I know what is the order $a^2$? –  Mary Jan 21 '13 at 20:03
    
@Mary: I've edited my answer so hopefully it is more clear, now. –  Clayton Jan 21 '13 at 20:13

Assume there exists exactly one and (one and only one ) element $a\in G$ of order $3$. Then the order of $|G| \ge 3$, and more importantly, there exists a subgroup $H = \langle a \rangle \leq G$ such that the order of $H$ = $|H| = |\langle a\rangle| = 3$. i.e., $a^3 = e$, the identity element of $G$. So $\langle a \rangle = \{e = a^3, a, a^2\}$.

Claim: if $a$ generates $H$, and the order of $a$ = |H| = 3, then $a^2 \neq a$ also generates $H$ and hence, $a^2$ is also of order $3$.

Recall, that the order of any cyclic subgroup (like $H = \langle a \rangle$) which is generated by the element $a$ is the order of the element that generates it. And the order of an element $a$ that generates a group $H$ is equal to the order of the group it generates. So we claim that $|\langle a^2 \rangle| = |H| = 3.$

We can see this more clearly by computing the elements of $\langle a^2\rangle$:

$a^2 \in \langle a\rangle,\;$ $(a^2)^2 = a^4 = a^3\cdot a = ea = a \in \langle a \rangle,\;$ and $(a^2)^3 = a^6 = (a^3)^2 = e^2 = e \in \langle a \rangle$.

Hence, $\langle a \rangle = \langle a^2 \rangle$. But then both $a$ and $a^2$ have order three, knowing that $a \neq a^2$, we have a contradiction, since we assumed $a$ was unique.

Hence there cannot be a unique (there cannot be one and only one) element of order $3$ in any group; if a group has an element of order $3$, it has at least two elements of order $3$.

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Deserves a positive! +1 –  Amzoti May 9 '13 at 2:50

Another approach to this: let $G$ be a group and $a \in G$ the only element of (finite) order $n$. Then $n=1$ or $2$. Proof : $ord(a)=ord(a^{-1}$), hence $a=a^{-1}$, that is $a^2=e$, the identity element. We conclude that $a$ is the identity element or $a$ has order 2.

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