# Finding the number of elements of order $2$ in a given group

How many elements of order $2$ are there in the group of order $16$ generated by $a$ and $b$ such that $o(a)=8$ and $o(b)=2$ and $bab^{-1}=a^{-1}$?

The basic thing i do not understand is that order of the generator must be equal to the order of the group. So how can $a$ and $b$ in this case generate the given group ?

I tried something like this, as $b$ is of order $2$ it must be its own inverse. So from the given condition on $a$ and $b$ replacing $b$ by $b^{-1}$ i get that $o(ba)=2$ . SO there are at least two elements of order two. However, this answer is obviously not good enough.

• The group you're considering is not cyclic; not even abelian, actually. – egreg Oct 3 '13 at 12:31
• You're missing the one given by a alone. If a has order 8, a^4 has order 2 – Chris Bonnell Oct 3 '13 at 12:32
• @AmanMittal: A cyclic group is one that is generated by one element. Your group is generated by two element, which is something different. – hmakholm left over Monica Oct 3 '13 at 12:33
• @AmanMittal: Yes, a cyclic group can have several possible generators. But each of these generators will generate the group alone. In your case neither $a$ nor $b$ generate the group alone; you need to use both before the group is generated. – hmakholm left over Monica Oct 3 '13 at 12:40
• @AmanMittal, your group can be given as $$G=\langle\;a,b\;;\;a^8=b^2=1\;,\;bab=a^{-1}\;\rangle$$ This is one member of the famous family of dihedral groups, namely $\;D_8\;$ . Perhaps you should google it... – DonAntonio Oct 3 '13 at 12:47

Hint: Consider $ba^{j}ba^{j}$ for $0 \leq j \leq 7.$
It's a group, so there's some identity element $1$. The order of $a$ is $8$, so $a,a2,…,a8=1$ are all distinct. The order of $a^2$ is $4$ since $1=a^8=(a^2)^4$ but $a^4$ and $a^6$ are not $1$. Since the order of $b$ is $2$, $b≠a$ and $b≠a^2$. Continue this process and work out a multiplication table.
$a^4;b;ba;ba^4$ are the four members of order $2$ of the given group given group $G=\{e,a,a^2,a^3,a^4,a^5,a^6,a^7,b,ba,ba^2,ba^3,ba^4,ba^5,ba^6,ba^7\}$