# Solving a Group theory problem using Cayley diagrams

• patric44
In summary, if G is a group and a,b belongs to G and O(a) = e , b.a =a.b^2 then find O(b) . However, this problem may be easier to solve using Cayley diagrams.f

#### patric44

Homework Statement
if G is a group and a,b belongs to G and O(a) = e , b.a =a.b^2 then find O(b) ?
Relevant Equations
b.a = a.b^2
hi guys
i saw this problem : if G is a group and a,b belongs to G and O(a) = e , b.a =a.b^2 then find O(b) , but i want to tackle this problem using Cayley diagrams , so my attempt is as following :
$$ba =ab^{2}$$
then i might assume b as flipping , a as rotation :
$$fr = rf^{2}$$
then knowing that ##r^{5} = e ## i suspect that the symmetry might be associated with a pentagon , but then i am stick here because i can't figure out the other substructure associated with this ##f^{n}## flipping .

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Homework Statement:: if G is a group and a,b belongs to G and O(a) = e , b.a =a.b^2 then find O(b) ?
What do O(a) and O(b) mean? I looked in my Abstract Algebra (Fraleigh) textbook, and wasn't able to find this notation. I'm tempted to think it means "order" but that doesn't seem to fit what you wrote.
patric44 said:
then knowing that ##r^5=e##
How do you know this. You don't show it as given information.

What do O(a) and O(b) mean? I looked in my Abstract Algebra (Fraleigh) textbook, and wasn't able to find this notation. I'm tempted to think it means "order" but that doesn't seem to fit what you wrote.

How do you know this. You don't show it as given information.
sorry i must have explained it more , in the problem it was given that ##O(a) = 5## so i suspected that a is a generator in which ## a^{5} = e##

Homework Statement:: if G is a group and a,b belongs to G and O(a) = e , b.a =a.b^2 then find O(b) ?
Relevant Equations:: b.a = a.b^2

hi guys
i saw this problem : if G is a group and a,b belongs to G and O(a) = e , b.a =a.b^2 then find O(b) , but i want to tackle this problem using Cayley diagrams , so my attempt is as following :
ba=ab2
then i might assume b as flipping , a as rotation :
fr=rf2
then knowing that r5=e i suspect that the symmetry might be associated with a pentagon , but then i am stick here because i can't figure out the other substructure associated with this fn flipping .
If your thought about looking at this problem as symmetries of a pentagon (which may or may not be the way to go), it might be helpful to look at the rotations and flips in terms of permutations of the vertices of the pentagon, with the vertices of the pentagon labelled as 1, 2, 3, 4, and 5. For example, here are two rotation permutations:
##r_0=\begin{pmatrix}1&2&3&4&5 \\ 1& 2 & 3 & 4 &5 \end{pmatrix}##
##r_1=\begin{pmatrix}1&2&3&4&5 \\ 2&3&4&5&1 \end{pmatrix}##
For ##r_0## each vertex is not rotated at all. For##r_1## vertex 1 is rotated to vertex 2, and so on, with each vertex moving to the next higher number.
For ##r_2##, each vertex is rotated by two positions
There are two more rotations.

For the flips, I count five different flips. For each one, one of the vertices is held fixed, and the other four vertices flip across the pentagon to the opposite. Each of the five flips can be represented by a permutation.
For example, ##f_1## holds vertex one constant.
##f_1=\begin{pmatrix}1&2&3&4&5 \\1&5&4&3&2\end{pmatrix}##
The other four flip permutations are similar.
If you make a Cayley diagram, it will need ten rows and ten columns, one row and column for each of the permutations. Possibly the diagram will help you answer the question of this problem, finding ##\mathcal O(b)##

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• patric44
The group $G$ is not $D_{10}$; if it were, the commutation relation would be $ba = a^{4}b$ and $b$ would be of order 2.

If $b = e$ then $G$ is $Z_5$, generated by $a$. I assume this is not the solution you're looking for.
If $b^2 = e$ then $ba = a$ and again $b$ is the identity.

So the order of $b$ is at least 3.

• patric44
Homework Statement:: if G is a group and a,b belongs to G and O(a) = e , b.a =a.b^2 then find O(b) ?
Relevant Equations:: b.a = a.b^2

hi guys
i saw this problem : if G is a group and a,b belongs to G and O(a) = e , b.a =a.b^2 then find O(b) , but i want to tackle this problem using Cayley diagrams , so my attempt is as following :
$$ba =ab^{2}$$
then i might assume b as flipping , a as rotation :
$$fr = rf^{2}$$
then knowing that ##r^{5} = e ## i suspect that the symmetry might be associated with a pentagon , but then i am stick here because i can't figure out the other substructure associated with this ##f^{n}## flipping .

I was able to construct $G$ as a semidirect product of the cyclic groups $Z_5$ and $Z_n$ where $n$ is the order of $b$. From $$ba = ab^2$$ I was able to obtain the conjugates $$bab^{-1} = ab, \qquad aba^{-1} = a^2 b^2 a^3 = b^{16}$$ from which I concluded that $Z_5$ must act on $Z_n$ rather than vice-versa. It follows from this that $Z_5$ is not acting on vertices of a pentagon, but on the group of rotations of an $n$-gon.

• patric44
I was able to construct $G$ as a semidirect product of the cyclic groups $Z_5$ and $Z_n$ where $n$ is the order of $b$. From $$ba = ab^2$$ I was able to obtain the conjugates $$bab^{-1} = ab, \qquad aba^{-1} = a^2 b^2 a^3 = b^{16}$$ from which I concluded that $Z_5$ must act on $Z_n$ rather than vice-versa. It follows from this that $Z_5$ is not acting on vertices of a pentagon, but on the group of rotations of an $n$-gon.
it turned out to more complicated than i thought using Cayley's diagram , so i started to do some factorization of conjugate relations and some right - left multiplications it took me a while but i guess i got it right , i got ##O(b) = 31## .