lal means the order of a(adsbygoogle = window.adsbygoogle || []).push({});

Theorem.

Let G be a group and a[tex]\in[/tex]G. Then lal=la^{-1}l

Proof.

Case 1, if lal=1

Case 2, if lal=n

Case 3, if lal=infinity

i understand case 1 and 3, so i'll be post the proof when need,

but, case 2

Here's the proof

Suppose lal=n

Then a^{n}=1 and a^{r}[tex]\neq[/tex]1, 1[tex]\leq[/tex]r<n-----------(1)

To show that (a^{-1})^{n}=1 and (a^{-1})^{r}[tex]\neq[/tex]1, 1[tex]\leq[/tex]r<n.

Clearly, (a^{-1})^{n}=(a^{n})^{-1}=(1)^{-1}=1

Suppose (a^{-1})^{r}=1, for some 1[tex]\leq[/tex]r<n

=> (a^{r})^{-1}, 1[tex]\leq[/tex]r<n

=> a^{r}=1, for some 1[tex]\leq[/tex]r<n

but this contradict (1)

So, (a^{-1})^{r}[tex]\neq[/tex]1, 1[tex]\leq[/tex]r<n

Hence, lal=n=la^{-1}l

I don't understand why, it suppose "(a^{-1})^{r}=1, for some 1[tex]\leq[/tex]r<n"

then say it contradict with (1), i cannot see how they contradict.

Help, T_T

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# Homework Help: Proof by contradicton

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