## Found Identity – Order of Group Elements

May 18, 2009 14:34 by scibuff

Another one in my series of Yahoo Answers this time using elementary properties of Groups.

Problem: Let $a,b\ \epsilon\ G$ be elements of group such that $(ab)^{k}=e$ where $e$ is the identity element of $G$. Prove that $(ba)^{k}=e$

Solution: The problem is indeed trivial if $G$ is abelian, i.e. $ab=ba$ for all $a,b\ \epsilon\ G$. One must be careful not to assume such properties when dealing with constructs in abstract algebra. Nevertheless, the solution below is quite simple and uses only associativity of Groups, i.e. $(ab)c=a(bc)$ for all $a,b,c\ \epsilon\ G$.

Lemma: For any $a,b\ \epsilon\ G$ there is a unique $x\ \epsilon\ G$ such that $ax=b$.

Proof: The lemma can be proven in two steps. First, show that such $x$ exists, and second, show that it must be unique.

Let $x=a^{-1}b$. Since $a,b\ \epsilon\ G$ and $G$ is a group, it follows from the existence ofÂ  inverses that $a^{-1}\ \epsilon\ G$ and from group closure $a^{-1}b\ \epsilon\ G$. Thus, $x\ \epsilon\ G$ (and the first step is done).

Let $x\ \epsilon\ G$ be such that $ax=b$, take $e$ to be the identity element of $G$. Then

$x=ex=(a^{-1}a)x=a^{-1}(ax)=a^{-1}b$

Thus it follows that if $ax=b$ then it must be that $x=a^{-1}b$, finishing the second step, hence proving out lemma.

Now we go back to our proof. Suppose

$(ab)^{k}=e$

is the identity in $G$, i.e.

$\overset{k\ times}{\overbrace{(ab)(ab)\cdots (ab)}}=e\ (1)$

Using group associativity we get

$a\overset{k-1\ times}{\overbrace{(ba)(ba)\cdots (ba)}}b=a(ba)^{k-1}b=e\ (2)$

Multiplying$(2)$ by $a$ (on right) we get

$a(ba)^{k-1}ba=a(ba)^{k}=ea$

$a(ba)^{k}=a\ (3)$

Now multiplying $(2)$ by $b$ (on left) we get

$ba(ba)^{k-1}b=(ba)^{k}b=be$

$(ba)^{k}b=b\ (4)$

Now from $(3)$ we have

$ba(ba)^{k}=ba\ (5)$

and from $(4)$

$(ba)^{k}ba=ba\ (6)$

Combining $(5)$ and $(6)$

$ba(ba)^{k}=(ba)^{k}ba=ba\ (7)$

Since $e$ is the identity element of $G$, by definition

$(ba)e=e(ba)=ba\ (8)$

Finally, using our Lemma and $(7)$ with $(8)$ it follows that $(ba)^{k}=e$