isomorphism problem

Apr 2010
22
0
Show that <a, b|a^5=b^2=e, ba=(a^2)b> is isomorphic to Z2.
 
Oct 2009
4,261
1,836
Show that <a, b|a^5=b^2=e, ba=(a^2)b> is isomorphic to Z2.
Let the game begin...(Happy) :

\(\displaystyle bab^{-1}=a^2\Longrightarrow ba^2b^{-1}=\left(bab^{-1}\right)^2=a^4\) , but also \(\displaystyle ba^2b^{-1}=b\left(bab^{-1}\right)b^{-1}=b^2ab^{-2}=a\) , so we get \(\displaystyle a=a^4\Longrightarrow a^3=1\) , which together with the given data \(\displaystyle a^5=1\)

means that \(\displaystyle a=1\) ,and thus in fact \(\displaystyle \left<a,b\;/\;a^5=b^2=1\,,\,ba=a^2b\right>=\left<b\;/\;b^2=1\right>\cong\mathbb{Z}_2\)

Tonio
 
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Apr 2010
22
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b(a^2)b^-1= (bab^-1)^2 due to that bab^-1 is a conjugate, and this is the same for ba^2b^-1= b(bab^-1)b^-1 is because of the definition of conjugate as well, right?
 
May 2009
1,176
412
b(a^2)b^-1= (bab^-1)^2 due to that bab^-1 is a conjugate, and this is the same for ba^2b^-1= b(bab^-1)b^-1 is because of the definition of conjugate as well, right?
The first part of your statement/question is right, but the second part is not true `because of the definition of conjugation'. Tonio has substituted in \(\displaystyle a^2 = bab^{-1}\) which can be done as \(\displaystyle ba=a^2b \Rightarrow bab^{-1} = a^2\).

How about...\(\displaystyle baba=a^2a=a^3\) (as \(\displaystyle bab = a^2\) by \(\displaystyle ba=a^2b\)). This gives us that \(\displaystyle bab=1\) by cancelling the \(\displaystyle a\)s and so we have that \(\displaystyle <a, b; a^5 = b^2=1, ba=a^2b> = <b; b^2>\) as required.