# isomorphism problem

#### aabsdr

Show that <a, b|a^5=b^2=e, ba=(a^2)b> is isomorphic to Z2.

#### tonio

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

• aabsdr

#### aabsdr

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?

#### Swlabr

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.