Suppose a^5 = e, b != e, and b^2 = aba(-1). Prove that

[dahmay]

Let G be a group, and let 'a' and 'b' be two elements of G.

Suppose that a^5 = e, where e is the identity, and suppose that b is not the identity. Suppose also that b^2 = aba^(-1).

Prove that lbl = 31

 

[stapel]

I'm sorry, but I do not understand your 'prove' statement. How does the element 'l' relate to the group and its other elements? And, since you're dealing in algebraic elements, where is the numerical value '31' coming in?

Please reply with clarification. Thank you.

 

[dahmay]

lbl=31 means the degree of element b equal thirty-one.
so the question show that lbl=31?
the subject about 'elments degree in groups'..

 

[stapel]

Please reply showing everything you have tried so far, and where you are stuck. Thank you.

Eliz.

 

[daon]

\(\displaystyle b^4=(b^2)^2 = (aba^{-1})(aba^{-1}) = ab^2a^{-1}\). So, \(\displaystyle b^4=ab^2a^{-1}\).

Continuing in this manner (THIS SHOULD BE PROVED WITH A LEMMA), we see that \(\displaystyle b^{2k}=ab^{k}a^{-1}\).

You are looking for the smallest such integer C such that \(\displaystyle b^C = e\).

So, notice \(\displaystyle 2^5 = 32\), and
\(\displaystyle b^{32} = ab^{16}a^{-1} = a(ab^{8}a^{-1})a^{-1} \\ \,\, = a^2(ab^4a^{-1})a^{-2} = a^3(ab^2a^{-1})a^{-3} \\ \,\, = a^4(aba^{-1})a^{-4} =a^{-5}ba^5\)

BUT! \(\displaystyle a^5 = e = a^{-5}\)

So,
\(\displaystyle b^{32} = a^{-5}ba^5 = b \\
\Leftrightarrow b^{32} = b \\
\Leftrightarrow b^{31}=e\)

Since 31 is prime, \(\displaystyle b^{31}=e\) IMPLIES |b|=31.

Hope that helps,
-daon

 

[pka]

Daon, that is very very good; way to go!

 

[daon]

Thanks!

While doing the proof I also found that proving the lemma \(\displaystyle b^{2^k} = a^kba^{-k}\), will allow easy proof for the following:

If p is prime and 2^p-1 is prime, with |a|=p and b is not the identity and you are given \(\displaystyle b^2 = aba^{-1}\) then |b| = 2^{p}-1.

 

[dahmay]

thak you for your help daon...