# Finding the elements of a group given two generators and relations

Hey everyone

Let's say I have two generators, $a$ and $b$, with the following relations:
$a^{5}=b^{2}=E$
$bab^{-1}=a^{-1}$;

Where E is the Identity element.

What I've done so far is this - the number of elements of the group is the product of the exponents of both generators, which is 10. Then I listed all the elements and their products which aren't equal to Identity. So I got the group elements:

{${E , a , a^{2} , a^{3} , a^{4} , ab , a^{2}b , a^{3}b , a^{4}b , b}$}

So is that right? The only thing that's confusing me is the second relation; not sure why its there really if u can get the group elements without it.

Thanks guys!

Staff Emeritus
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You are missing a ton of elements - a and b do not commute, so a2b is not the same as aba for example. Basically if all you had was a5=b2 = E, then you would have everything of the form
ak1bak2bak3b....aknb for any choice of k1,...,kn (and number n) as distinct elements. The second relation will make some of them equal to each other.

okay, so I have two questions:

(1) How do u know that a and b don't commute? I mean I couldnt infer that from the context of the question
(2) do I just need elements like ba, ba^2, ba^3 etc...since ab is not = ba?

I always thought that the order of the group is the product of the powers of its generators - so that's not generally true then?

Staff Emeritus
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When they list relations, those are the ONLY relations that a and b satisfy (along with anything that can be generated by stringing together those relations). a and b commuting is a relation:
ab = ba.

If they don't tell you that, or things from which you can infer this, then it's not true. And yes, you will also need elements that start with b (and elements that start and end with b, and elements that start and end with a).

Your statement about the order of a group is only true for an abelian group, if I give you a group generated by a and by with a2 = b2 = E and no other relations, this group is infinitely large, with elements a, ab, aba, abab, ababa,.... and b, ba, bab, baba, etc.

Okay, thanks for the explanation :) but....where do I stop with writing the elements?

Staff Emeritus
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If there are infinitely many elements in a set, then you just can't write them all down. If the relations happen to make your set finite, then you stop when you've written them all down...

You have to look at that extra relation that you have and determine how it lets you re-write some of these elements.... as a warmup, how would you use it to re-write ba in terms of a and b only and in that order (a's, then b's)) (no inverses)

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I've been trying for ages and I don't know how i can re-write ba in terms of a and b....I keep getting random stuff and I dont even know if its right. Is this an infinite group...?

Just to make sure...I have the relation
$bab^{-1}=a^{-1}$
So if I multiply both sides by, for example $b^{-1}$, do I get this:
$b^{-1}bab^{-1}=b^{-1}a^{-1}$
$ab^{-1}=b^{-1}a^{-1}$

So basically is the order of my operations correct?

I mean I have a lot of questions about how these operations are meant to be...a lot of stuff doesnt make sense, cos unless im wrong I just showed that b is its own inverse...so something's not right

Staff Emeritus
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OK, it might help to start even more basic. Given that b2 = E, what is b-1 in terms of b? Can you calculate a similarly?

Well it depends...cos if what im doing is wrong then that might be the problem. Check this out:
$b^{2}=E$ Good so far
$b^{2}.b^{-1}=E.b^{-1}$
$b=b^{-1}$

But..that's wrong isnt it :S

and as far as $a$ goes, I'm not sure if I can do any of the regular power manipulation with it, cos I dont know the composition rule of the group. Can I still assume that, for example, the square root of $a^{5}$ is basically $a^{5/2}$?

I've managed to derive a handful of other relations, which are eliminating elements of the form aba, aba^2 etc from the group......if that's correct, I think ive reached a maximum number of 14 elements in this group, all other things like ababa seem to be simplifying to preceding elements

I do wana thank you for your time though, i really really appreciate it. You're helping me a lot :D

Okay, this is where I've reached.
I used the relation $bab^{-1}=a^{-1}$ to show that
$aba=b$

Now, if that is correct (please tell me if it is) then I cant find any more unique elements than the following ones

${E, a, a^{2}, a^{3}, a^{4}, b, ba, ba^{2}, ba^{3}, ba^{4}, ab, ab^{2}, ab^{3}, ab^{4}}$

which has an order of 14.

Staff Emeritus
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Well it depends...cos if what im doing is wrong then that might be the problem. Check this out:
$b^{2}=E$ Good so far
$b^{2}.b^{-1}=E.b^{-1}$
$b=b^{-1}$

But..that's wrong isnt it :S

and as far as $a$ goes, I'm not sure if I can do any of the regular power manipulation with it, cos I dont know the composition rule of the group. Can I still assume that, for example, the square root of $a^{5}$ is basically $a^{5/2}$?

This looks perfectly good for b... can't you do something similar for a?

a5 = E
a5 a-1 = a-1
a4 = a-1.

Then bab-1 = a-1 turns into
ba = a-1b = a4 b.

So this basically tells you how to "commute" a and b, which is another way to approach the problem. However your way also works (modulo a couple mistakes I think you made), once we have
aba = b

then we know that any element which has an a to both the right and the left of b we can reduce using this. Therefore the only possible elements are
biakbj

where i=0 or 1, k = 0,1,2,3,4 and j=0 or 1

again it's critical here that a-1 and b-1 can be written in terms of a and b respectively, otherwise we would have elements that have a bunch of inverses in them as well. Over all choices of i,j and k there are 20 possible elements, and I think they're all distinct but I could be wrong.

But I don't need to have inverses in the group right? T_T

I understand this a ton better now. Watch me show off to my friends Y'ALL ARE WRONG LOLZ

ahem. I just woke up, excuse me

economicsnerd
Most of this is said above, but to synthesize:
- $a^{-1}=a^4$ and $b^{-1}=b$, so that every element can be expressed using only nonnegative powers.
- $aba=b$, and so any element can be expressed in the form $b^ia^jb^k$ for some $i,j,k\geq 0$.
- $a^5=b^2=e$, and so we can always take $i,k\in \{0,1\}, j\in\{0,1,2,3,4\}$.
- If $j=0$, then there's further redundancy, as $b^ib^k$ depends only on whether $i+k$ is even or odd (given $b^2=e$).
Therefore we can write $G=\{b^ia^jb^k: \enspace i=0,1, \enspace j=1,2,3,4, \enspace k=0,1 \}\cup\{e,b\}$.

I think there's no redundancy here, so that $|G|= 2\cdot 4\cdot 2 + 2 = 18.$

Homework Helper
This looks perfectly good for b... can't you do something similar for a?

a5 = E
a5 a-1 = a-1
a4 = a-1.

Then bab-1 = a-1 turns into
ba = a-1b = a4 b.

So this basically tells you how to "commute" a and b, which is another way to approach the problem. However your way also works (modulo a couple mistakes I think you made), once we have
aba = b

then we know that any element which has an a to both the right and the left of b we can reduce using this. Therefore the only possible elements are
biakbj

where i=0 or 1, k = 0,1,2,3,4 and j=0 or 1

again it's critical here that a-1 and b-1 can be written in terms of a and b respectively, otherwise we would have elements that have a bunch of inverses in them as well. Over all choices of i,j and k there are 20 possible elements, and I think they're all distinct but I could be wrong.

There are 10 elements; the group is $D_5$, the symmetry group of the regular pentagon.

Briefly, you have $ba = a^4b$, from which it follows that $ba^k = a^{5-k}b$ for $k \in \{1,2,3,4,5\}$. So every element is expressible in the form $a^jb^k$ and the 10 distinct elements are $\{e, a, a^2, a^3, a^4, b, ab, a^2b, a^3b, a^4b\}$.

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Homework Helper
Most of this is said above, but to synthesize:
- $a^{-1}=a^4$ and $b^{-1}=b$, so that every element can be expressed using only nonnegative powers.
- $aba=b$, and so any element can be expressed in the form $b^ia^jb^k$ for some $i,j,k\geq 0$.
- $a^5=b^2=e$, and so we can always take $i,k\in \{0,1\}, j\in\{0,1,2,3,4\}$.
- If $j=0$, then there's further redundancy, as $b^ib^k$ depends only on whether $i+k$ is even or odd (given $b^2=e$).
Therefore we can write $G=\{b^ia^jb^k: \enspace i=0,1, \enspace j=1,2,3,4, \enspace k=0,1 \}\cup\{e,b\}$.

I think there's no redundancy here, so that $|G|= 2\cdot 4\cdot 2 + 2 = 18.$

Lagrange's theorem and $a^5 = e$ say otherwise.

economicsnerd
Lagrange's theorem and $a^5 = e$ say otherwise.

You're right. That's embarrassing.

Okay....I'm a little confused now...

economicsnerd
Comment #20 is exactly correct.

Okay, cos I got 14 elements, but let me double check to see if they can be reduced to 10. I have another question, though - is this just a coincidence? I mean that the order of this group is 10 - the product of the powers of its generators - even though it's non-abelian?

While I'm at it let me just confirm a few results I got, cos if they are right then I think I got the hang of manipulating ordered products. So I got these relationships, could u tell me if they are right?
$aba = b$
$ab = ba^{4}$
$ba = a^{4}b$
$a^{2}b = aba^{4}$, which becomes, using the first equation, $a^{2}b=ba^{3}$

So my group has been reduced to 10 elements, which are $E, a, a^{2}, a^{3}, a^{4}, b, ab, ba, ba^{2}, ba^{3}$

NOTE - I have kept the elements of the form $ba^{k}$, but pasmith has used the other equality instead that $ba^{2} = a^{3}b$ etc.

Gold Member
First of all this group has 10 elements. Your first list was correct.

In your second list you forgot that b$^{2}$ = id.

A group with two sets of generators of finite order,n amd m, does not have to be abelian to have order nxm.

Whenever the cyclic subgroup generated by one is normal this will be true. See if you can prove this.

In your example the cyclic subgroup of order 5 generated by b is normal. This is what the second relation says.

The relation aba$^{-1}$ = b$^{-1}$ defines a group no matter what the order of b. If b has order 4 for instance it is the dihedral group of order 8.

Good examples of easy finite groups can be made by starting with two orthogonal matricies with only 1,-1, and 0 entries and taking all of the multiples. You will find many interesting groups this way.

For instance the dihedral group of order 8 is generated by the matricies

a =

0 -1
1 0

which a a positive rotation of 90 degrees

and

a =

1 0
0 -1

which is a reflection around the y axis.

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