- #1
annoymage
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lal means the order of a
Theorem.
Let G be a group and a[tex]\in[/tex]G. Then lal=la-1l
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 an=1 and ar[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=(an)-1=(1)-1=1
Suppose (a-1)r=1, for some 1[tex]\leq[/tex]r<n
=> (ar)-1, 1[tex]\leq[/tex]r<n
=> ar=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-1l
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
Theorem.
Let G be a group and a[tex]\in[/tex]G. Then lal=la-1l
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 an=1 and ar[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=(an)-1=(1)-1=1
Suppose (a-1)r=1, for some 1[tex]\leq[/tex]r<n
=> (ar)-1, 1[tex]\leq[/tex]r<n
=> ar=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-1l
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