# Non-Abelian Group of Order 21

1. Nov 12, 2012

### sammycaps

So I was working through some problems in Herstein's Algebra on my own time, and I came across something I wasn't so sure about.

The question was, Find a non-abelian group of order 21 (Hint: let a3=e and b7=e and find some i such that a-1ba=bi≠b which is consistent with the assumptions that a3=e and b7=e)

All the solutions say if we set i=2, then this generates a group of order 21. I was just wondering how exactly I could check this without using the semiproduct stuff, which I haven't yet learned. I understand that consistent means that it doesn't contradict the assumptions, but I'm exactly sure how to check that, and what constitutes a full checking that there is no contradiction. Taking a-1ba=bi≠b to all its powers and checking to see that they give no contradiction doesn't seem to be the right way to go about it.

Similarly, is there a quick way to say 3, 5 and 6 don't work (assuming I'm right that they don't)? I can check and find that they give a contradiction to the assumptions, but I don't know if this is the easiest way.

Last edited: Nov 12, 2012
2. Nov 12, 2012

### topspin1617

Well, semidirect product really is the right way to do it... the theory guarantees a unique (up to isomorphism) non-abelian group of order pq, whenever p<q are primes with p|(q-1).

Forgoing that, I guess direct calculations are the only way to go. I suppose you could write out a multiplication table for the group to see that everything works out.

There is a reason underlying this, but I'm not sure that just quoting it is really a proof that the construction works (i.e., that it's sufficient... it is if you go through the whole semidirect construction, but I think we can see this much without that): any identity of the form $a^{-1}ba=b^i$ induces an automorphism $\varphi_{a}:\langle b\rangle\rightarrow\langle b\rangle$ (because conjugation is always injective, and every power of $b$ generates$\langle b\rangle$ because 7 is prime. That is, the identity you write down induces a map $\left\langle a\right\rangle\rightarrow \mathrm{Aut}\left\langle b\right\rangle$ given by $a\mapsto \varphi_a$. Under any group homomorphism, an element of order 3 must map to an element of order 3 (or 1). You can check that, with i=2 above, conjugating b by a 3 times gives you b again; that is, the automorphism induced by a has order 3, which is good. If i=3, though, conjugating by a three times produces b^6. You can see it directly in the group:

$a^{-3}ba^3=b^6\Rightarrow b=b^6\Rightarrow b^5=e\Rightarrow b=e$.

This is really the core of the issue. Now, this really is the beginnings of semidirect product, and I know you haven't learned it yet, but I think it is possible to see what is going on in terms of conjugation inducing an automorphism of a (normal) subgroup.

3. Nov 13, 2012

### sammycaps

Thanks! But even if I were to write out the multiplication table, wouldn't I still need to do a bunch of algebraic manipulations to make sure I don't have a contradiction? And then, it would be difficult to be sure that I've exhausted all the possibilities, right? Maybe not, I'm not so sure.

I guess semidirect product is really the way to go. The explanation makes sense, I guess I'll just have to wait.

4. Nov 13, 2012

### lavinia

The semi direct product idea is worth learning. It gives a method of for constructing many groups.

here is maybe a slightly different tack in your case.

The element aba may be rewritten as aba$^{-1}$a$^{2}$

This means that the multiplication in the group can be defined on all pairs (b$^{i}$,a$^{j}$)

by the rule (b$^{i}$,a$^{j}$).(b$^{k}$,a$^{h}$) = (b$^{i}$.a$^{j}$b$^{k}$a$^{-j}$,a$^{j+h}$)

This rule is well defined as long as conjugation of <b> is an automorphism of order 3. If conjugation ,for instance, were the identity map (which is certainly of order 3) the group would be the direct product of the two cyclic groups. That is aba would equal ba$^{2}$.

Note that there are 21 pairs (b$^{i}$,a$^{j}$) so both groups are of order 21.

Here is another example. Let b be of order 4 and a of order 2. Set aba$^{-1}$ = b$^{3}$, This will give you a non-abelian group of order 8. If on the other hand if aba$^{-1}$ = 1 then the group is the product and is also of order 8.

Last edited: Nov 13, 2012
5. Nov 13, 2012

### sammycaps

Ok, I'm sorry if this sounds stupid, but why does this being well-defined necessarily mean that there is no contradiction? Also, does this not look something like the explanation topspin1617 gave where he said this was the beginning of semidirect product (I just started reading up on it, and it looks like it is, but I don't know)?

Also, can you suggest a reference online for this that I can look up?

I'm sorry if I sound a bit confused.

6. Nov 14, 2012

### lavinia

If it is well defined how do you think it could have a contradiction?
I tried to write things out so that you could see why it is called a semi direct product as opposed to a direct product.

I strongly suggest that you work through the formulation I gave in the case of your two cyclic groups.

Last edited: Nov 14, 2012
7. Nov 15, 2012

### mathwonk

i wonder if this can be realized as a sort of rotation subgroup of the group of collineations of the finite plane of 7 points and 7 lines. that group has 168 elements, and may have a subgroup of order 21. maybe the rotations can be made to act on the 7 points, with 3 rotations fixing each point. ??? just trying to think of some place where you might "see" this group geometrically.

bah..i don't see it. these rotations don't seem to form a group, (if they even exist).

Last edited: Nov 15, 2012
8. Nov 15, 2012

### lavinia

you could write down a matrix group simply. it would be generated by two 7x7 matrices. you could then generate the multiplication table in excel

9. Nov 15, 2012

### mathwonk

i think this is the projective plane over the field of 2 elements, so it is given fully by the invertible 3by3 matrices over Z/2??