Question about presentation of cyclic group

**tttcomrader** · September 15th 2008, 06:41 AM

I want to list the elements of $T=<a,b : a^6=1, b^2=a^3,bab^{-1}=a^{-1} >$

So from my understanding, $a^ia^j=a^{i+j} \ \ \ \forall 0 \leq i,j \leq 5$ with $a^{i+j}=a^{i+j-6} \ \ i+j \geq 6$

But I'm having trouble trying to make senses of the two other conditions, and how to properly translate them into the table.

Thank you!!!

**Opalg** · September 15th 2008, 08:22 AM

Notice first that $b^4 = (b^2)^2 = (a^3)^2 = a^6 = 1$ .

From the relation $bab^{-1} = a^{-1}$ it follows that $ba = a^{-1}b = a^5b$ . Using that, in any product of a's and b's you can always push the a's to the left of the b's. For example $aba^2b = a(ba)ab = a(a^5b)ab = bab = a^5b^2$ . In that way, you can express any element of the group as $a^ib^j$ , with $0\leqslant i\leqslant5$ and $0\leqslant j\leqslant3$ . That gives you 24 elements of the group, and you still have a fair amount of work to do if you want to write down the whole multiplication table.

**NonCommAlg** · September 15th 2008, 12:10 PM

and to complete Opalg's argument

: also note that since $b^2=a^3$ and $b^3=a^3b,$ every element of $T$ can be written as $a^ib^j$ with $0 \leq i \leq 5$ and $0 \leq j \leq 1.$ it's easy to show that this

presentation is unique. so $T$ is a group of order $12$ with elements $1,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b.$ the group $T$ is called dicyclic group.

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