Group theory (FP3)

The group is modulo 11, and you want subgroups of order 2 or 5, so find the numbers that are the multiplicative inverses of 2 or 5 in the group $\mathbb{Z}/11\mathbb{Z}$. The cyclic groups generated by these elements are the subgroups you want.

I'm sorry, I don't quite follow, could you perhaps explain in a simpler way? When you say find the "multiplicative inverses of 2 and 5" do you mean the inverse elements of 2 and 5 in the group M?

NOPE! I was wrong, I misread and thought it was an additive group.

In this case you want $x^2$ or $x^5$ equal to 1 mod 11 which is a far harder problem, so the best is probably to go with Cayley tables.

Sorry!

Ah, no the binary operation is multiplication mod 11. No worries! I'll just have to do it the old fashioned way.

Do you know anything that's special if the order of the group is prime?

EDIT: I'm not sure I understand your question, you've already answered it without needing to write out the tables!

I have already done part i) (By Lagrange's Theorem all proper subgroups have orders which are factors of the order of the group, 10 = 2x5 2 and 5 are both prime, groups of prime order are cyclic etc...) and part ii). I know how to do part iii) and I've done it correctly, I checked my answers with the mark scheme, but I had to write out the Cayley Table for the group to do this. I was just wondering if there was an easier way to find all the proper subgroups without having to write out the Cayley Table.

Can you show me the markscheme? (Which board and year is the paper from?) It seems tedious to not just be able to say "clearly M has subgroups of order 2 and 5, which have to be cyclic by (ii)" to finish off (iii)?

This is FP3 (OCR MEI) June 07. For part iii) you have to specifically list all the subgroups, with all their elements, you can't just say what orders they will have.

I suppose that's fair actually, otherwise the question would be very trivial. It's very easy to figure out what the groups must be.

Please enlighten me. We obviously have $\{1, 10\}$. How can we easily determine the elements of the subgroup of order 5?

Wait, you are allowed a calculator for this exam, right?

Yes.

Then actually it is quite easy. You just need to solve the equation $x^5 - 1 \equiv 0 \pmod{11}$.

Any easy way of doing part iii)? By Lagrange's theorem we must have subgroups of order 2 and 5, and by part i) these subgroups must be cyclic... is there any easy way to do this without writing out the Cayley table for the group?

For $x \in M$? So this will give you an element of order 5, which will be the generator for the cyclic subgroup of order 5,

Minor point (not worth worrying about if it distracts you from the exam's main content), but the existence of order-2 and order-5 subgroups is in fact a consequence of Cauchy's Theorem rather than Lagrange's. Lagrange tells you that all nontrivial proper subgroups are of order 2 or 5, but it doesn't tell you that such subgroups actually exist. Cauchy's Theorem gives you existence.

The solutions to this equation in this case is x = 1,3,4,5,9. Now x = 1 clearly must be omitted, but with any other x, yes a cyclic subgroup is generated as if $x^p = 1$ in any group where $p$ is prime and $x$ is not the identity, then the subgroup generated by $x$ MUST be cyclic (this follows from Lagrange's theorem).

I understand yeah, thanks for this. So to confirm (even though it's trivial) for the subgroup of order 2 we would be solving the equation:

$x^2 - 1 \equiv 0 \pmod{11}$?

Yep!