MHB Show that the elements have the same order

[mathmari]

Hey!

Show that the elements $ab$ and $ba$, with $a,b \in G$ have the same order.
The same stands also for the elements $a$ and $c^{-1}ac$.

I have done the following:

• Let $m=ord(ab) \Rightarrow (ab)^m=1 \Rightarrow ababab \dots abab=1 $
  
  Since $a \in G$, $a^{-1} \in G$:
  
  $\Rightarrow a^{-1}ababab \dots abab=a^{-1} \Rightarrow babab \dots abab=a^{-1} \\ \Rightarrow babab \dots ababa=a^{-1}a \Rightarrow babab \dots ababa=1$
  
  Since we have still the same number of elements ($2m$), we have that $(ba)^m=1$.
  
  Therefore, $ord(ab)=ord(ba)$.Is this correct?? (Wondering)

• Let $ord(a)=m \Rightarrow a^m=1 \tag 1$
  
  Let $ord(c^{-1}ac)=n \Rightarrow (c^{-1}ac)^n=1 \Rightarrow c^{-1}acc^{-1}ac \dots c^{-1}ac=1 \Rightarrow c^{-1}aa \dots ac=1$
  
  Since we multiplied $n$ times the element $c^{-1}ac$ at itself, we have that $c^{-1}a^nc=1 \Rightarrow cc^{-1}a^nc=c \Rightarrow a^nc=c \Rightarrow a^ncc^{-1}=c c^{-1} \Rightarrow a^n=1 \tag 2$
  
  From $(1)$ and $(2)$ we have that $m \mid n$, right ?? But how could I continue to show that $m=n$?? (Wondering)

 

[I like Serena]

$\DeclareMathOperator\ord{ord}$
Hey! :)

Since we have still the same number of elements ($2m$), we have that $(ba)^m=1$.

Therefore, $\ord(ab)=\ord(ba)$.Is this correct?? (Wondering)

You have proven $\ord(ab)=m \Rightarrow (ba)^m=1$.
That only means that $\ord(ba)|m$.

What more do you need to get $\ord(ab)=\ord(ba)$? (Wondering)

Let $ord(a)=m \Rightarrow a^m=1 \tag 1$

Let $ord(c^{-1}ac)=n \Rightarrow (c^{-1}ac)^n=1 \Rightarrow c^{-1}acc^{-1}ac \dots c^{-1}ac=1 \Rightarrow c^{-1}aa \dots ac=1$

Since we multiplied $n$ times the element $c^{-1}ac$ at itself, we have that $c^{-1}a^nc=1 \Rightarrow cc^{-1}a^nc=c \Rightarrow a^nc=c \Rightarrow a^ncc^{-1}=c c^{-1} \Rightarrow a^n=1 \tag 2$

From $(1)$ and $(2)$ we have that $m \mid n$, right ?? But how could I continue to show that $m=n$?? (Wondering)

You have effectively started from $\ord(c^{-1}ac)=n$ to show that $\ord(a)|n$.
Suppose you start from $\ord(a) = m$, what does that mean for $\ord(c^{-1}ac)$? (Wondering)

 

[mathmari]

$\DeclareMathOperator\ord{ord}$
You have proven $\ord(ab)=m \Rightarrow (ba)^m=1$.
That only means that $\ord(ba)|m$.

What more do you need to get $\ord(ab)=\ord(ba)$? (Wondering)

$m=\ord(ab), n=\ord(ba)$

$(ab)^m=1 \Rightarrow (ba)^m=1 \Rightarrow n \mid m$

$(ba)^n=1 \Rightarrow bababa \dots baba=1 \Rightarrow bababa \dots baba=a^{-1}=a^{-1} \Rightarrow bababa \dots bab=a^{-1} \\ \Rightarrow abababa \dots bab=aa^{-1} \Rightarrow abababa \dots bab=1 \Rightarrow (ab)^n=1 \Rightarrow m \mid n$

From $n \mid m$ and $m \mid n$ we have that $n=cm$, right?? How can I show that $c=1$ such that $n=m$?? (Wondering)

Is the way I show it correct or is there a better way to prove it?? (Malthe)

You have effectively started from $\ord(c^{-1}ac)=n$ to show that $\ord(a)|n$.
Suppose you start from $\ord(a) = m$, what does that mean for $\ord(c^{-1}ac)$? (Wondering)

I have to show also that $n \mid m$, right??

$a^m=1 \Rightarrow aa \dots aa=1 \Rightarrow acc^{-1}acc^{-1}a \dots cc^{-1}acc^{-1}a=1=cc^{-1} \\ \Rightarrow c^{-1}acc^{-1}acc^{-1}a \dots cc^{-1}acc^{-1}ac=c^{-1}cc^{-1}c \Rightarrow c^{-1}acc^{-1}acc^{-1}a \dots cc^{-1}acc^{-1}ac=1$

Are there $m$ terms of $c^{-1}ac$ ?? (Wondering)

 

[I like Serena]

$\DeclareMathOperator\ord{ord}$

$m=\ord(ab), n=\ord(ba)$

$(ab)^m=1 \Rightarrow (ba)^m=1 \Rightarrow n \mid m$

$(ba)^n=1 \Rightarrow bababa \dots baba=1 \Rightarrow bababa \dots baba=a^{-1}=a^{-1} \Rightarrow bababa \dots bab=a^{-1} \\ \Rightarrow abababa \dots bab=aa^{-1} \Rightarrow abababa \dots bab=1 \Rightarrow (ab)^n=1 \Rightarrow m \mid n$

From $n \mid m$ and $m \mid n$ we have that $n=cm$, right?? How can I show that $c=1$ such that $n=m$?? (Wondering)

Is the way I show it correct or is there a better way to prove it?? (Malthe)

We have:
$$n \mid m \Rightarrow n\le m$$
$$m \mid n \Rightarrow m\le n$$
It follows that:
$$m=n$$
(Emo)

I have to show also that $n \mid m$, right??

$a^m=1 \Rightarrow aa \dots aa=1 \Rightarrow acc^{-1}acc^{-1}a \dots cc^{-1}acc^{-1}a=1=cc^{-1} \\ \Rightarrow c^{-1}acc^{-1}acc^{-1}a \dots cc^{-1}acc^{-1}ac=c^{-1}cc^{-1}c \Rightarrow c^{-1}acc^{-1}acc^{-1}a \dots cc^{-1}acc^{-1}ac=1$

Are there $m$ terms of $c^{-1}ac$ ?? (Wondering)

Yep! (Mmm)

 

[mathmari]

We have:
$$n \mid m \Rightarrow n\le m$$
$$m \mid n \Rightarrow m\le n$$
It follows that:
$$m=n$$
(Emo)

Ahaa.. Ok! (Smile)

Yep! (Mmm)

So we have $(c^{-1}ac)^m=1$.
$\Rightarrow n \mid m$

Therefore,
$$n \mid m \Rightarrow n\le m$$
$$m \mid n \Rightarrow m\le n$$

$$ \Rightarrow m=n$$
Is it the only way to prove the sentence?? (Wondering)

 

[Deveno]

I find it easier to use this fact, for any $m \geq 0$:

$(ba)^{m+1} = b(ab)^ma$ (can you prove this using induction?).

Now if $(ab)^m = 1$, then:

$(ba)^{m+1} = b(ab)^ma = ba$.

Multiplying both sides by $a^{-1}b^{-1}$ we get:

$(ba)^m = 1$.

Suppose that for some $0 < k < m$, we have:

$(ba)^k = 1$.

It follows that:

$a(ba)^kb = ab$, but the LHS is: $(ab)^{k+1}$, so this is:

$(ab)^{k+1} = ab$, and multiplying by $b^{-1}a^{-1}$ on both sides, we have:

$(ab)^k = 1$. This is a contradiction, since $0 < k < m$, and $m$ is the SMALLEST positive integer for which:

$(ab)^m = 1$.

So we know that:

(1) the order of $ba$ divides $m$.
(2) the order of $ba$ is not less than $m$.

The only positive integer which qualifies is $m$.

It is also handy to remember that:

$(cac^{-1})^k = ca^kc^{-1}$ (this can also be proved by induction).

So if the order of $a$ is $m$, then:

$(cac^{-1})^m = ca^mc^{-1} = cc^{-1} = 1$.

So the order of $cac^{-1}$ divides $m$, so is at MOST $m$.

Again, suppose that $(cac^{-1})^k = 1$, for some $0 < k < m$.

Then $ca^kc^{-1} = 1$ so that:

$a^kc^{-1} = c^{-1}$

$a^k = c^{-1}c = 1$, which is impossible.

***************************

In other words, to show $m = n$ when $n|m$, we can show that $m|n$, OR:

we can show that $0 < n < m$ is not possible.