Order of group, Order of element

[LagrangeEuler]

If group $(G,\cdot)$ is defined with two generators $a$ and $b$. And $a^n=e$, $b^{m}=e$. Is there any Theorem to tell us what is the largest group they can form?

 

[member 587159]

Yes, the group $\langle a,b\mid a^n = e, b^m = e \rangle$, which is a quotient of the free group on $\{a,b\}$.

 

[fresh_42]

... and without any commutation relations, this is an infinite group.

 

[LagrangeEuler]

I do not understand. For instance let take example $a^2=e$, $b^3=3$, $aba=b$. Could you explain me on that example?

 

[fresh_42]

I do not understand. For instance let take example $a^2=e$, $b^3=3$, $aba=b$. Could you explain me on that example?

Sure. $aba=b \Longrightarrow ab=ba^{-1}=ba$ hence we have a commutative group. Now $\langle a \rangle \times \langle b \rangle = \mathbb{Z}_2\times \mathbb{Z}_3= \mathbb{Z}_6$.

 

[LagrangeEuler]

Great. Interesting. You easily find that. But to understand could we see just one more example $a^{4}=b^{4}=e$, $aba=b$? It should be noncommutative. But it is not easy to find order. Should it be at least 16?

 

[fresh_42]

Great. Interesting. You easily find that. But to understand could we see just one more example $a^{4}=b^{4}=e$, $aba=b$?

Looks like $Q_8=\langle x,y|x^4,x^2y^{-2},yxy^{-1}x\rangle$. Maybe I can figure it out - or you.

 

[pasmith]

If you have a commutativity relation, you can turn any string of $a$'s and $b$'s into a string of the form $a^kb^l$, where $0 \leq k < n$ and $0 \leq l < m$. There are exactly $nm$ such strings, and each such string gives a distinct element of the group. Thus the order of the group is $nm$.

 

[LagrangeEuler]

Looks like $Q_8=\langle x,y|x^4,x^2y^{-2},yxy^{-1}x\rangle$. Maybe I can figure it out - or you.

This is a mistake than in the link?
https://math.stackexchange.com/ques...yejr_s-kcorwvMyFwZUeVMcxO_0OMKGMFqBZBzW5c_t-w

"The smallest example occurs in order 16. There are 3 nonisomorphic groups of order 16, each having 3 elements of order 2 and 12 elements of order 4. One of them is the abelian group, direct sum of two cyclic groups of order 4. Another is the direct product of the quaternion group of order 8 with a cyclic group of order 2. The third one is generated by two elements $a$ and $b$ of order 4 with the relation $aba=b$.

In the book by Thomas and Wood, Group Tables, these are called 16/3, 16/7, and 16/10, respectively. I don't know whether other sources use this numbering."

 

[fresh_42]

Yes, you are right, there are $16$ elements, listed by @pasmith in post #8. Hence the task is to calculate the group table or list all subgroups to identify which of the nine non Abelian groups it is.

 

[LagrangeEuler]

Yes, you are right, there are $16$ elements, listed by @pasmith in post #8. Hence the task is to calculate the group table or list all subgroups to identify which of the nine non Abelian groups it is.

It is somehow too hard to see it sometimes from the relation between generators. From $a^4=e$,$b^4=e$ it should be at least group of order $8$ from Lagrange theorem. Right? However it could be $12$,$16$... It is very hard to see. :(

 

[fresh_42]

All elements $a^nb^m$ with $0\leq n,m <4$ are different, which are sixteen. If it was Abelian then we would get $a^2=1$ from $aba=b$, which we excluded. So we have nine non Abelian groups of order $16$. Next I would exclude elements of order eight, which reduces the possibilities to five groups.

 

[pasmith]

To construct the Cayley table, observe that if $aba = b$ then $ba = a^3 b$, from which it follows that $$b^k a^l = a^{3^kl}b^k$$ since moving an $a$ to the left of a $b$ gives you three times as many $a$'s. Hence $$(a^k b^l)(a^p b^q) = a^{k+3^l p}b^{l+q}.$$

 

[ChinleShale]

You might have fun playing around with pairs of signed permutation matrices with only 1,-1,and zero as entries. Start with arbitrary a and b with finite powers and see what groups they generate.

For instance $a= \begin{pmatrix}0&0&0&^{-}1\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}$ and $b=\begin{pmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{pmatrix}$

Here $a^8=1$ and $b^2=1$

Notice that the subgroup generated by $a^2$ and $b$ is the dihedral group of order 8