Groups: determining elements with polynomials

I had a problem that goes like this: I have the grop G = (Z₂₉\{0}, multiplication). It has exactly one subgroup H with four elements and exactly one subgroup K with 14 elements. The task is to detemine H ∩ K - this group has a size ≠ 1.

We know the identity element is part of both, H and K wouldn't be groups else. Lagrange's theorem gives us that the size must be two.

Now, in the solution they do this to get the second element: the second element has an order that divides the number of elements in the group. Question #1: how do they know this? Is it perhaps a theorem that I've missed? Question #2: how can an element have an order? I know groups can have it (it's just the number of elements in the group), and permutations can have it (number of times you have to permute to get to start again), but a single element seems a bit counterintuitive.

From this they draw the conclusion that the element must solve the equation x² - 1 = 0. Question #3: how do they get this equation? This "order" of the element were another number, say a, would I have to make an a:th degree polynomial, and how would I get the coefficients and other numbers? Besides, how do they get the - 1? Is that the identity element?

Grateful for answers!

The order of an element x is the least positive integer k such that x^k= the identity.

If you know that there are just two elements, x and the identity, then you must have x^2=identity or as they have it x^2 - 1 =0.

Thank you a lot!

the order of an element in a group, is the number of times you have to "multiply" (multiply is in quotes because i am referring to the group operation which may not actually be ordinary multiplication) it by itself to reach the identity. if a group is finite, this is always a finite positive integer (all the powers of an element x can't be different, so two of them must be the same; in which case, if r > s, then x^r-s =e, the identity of the group).

Lagrange's theorem says that the order of a subgroup must divide the order of a group. since the cyclic subgroup generated by x, <x> = {x,x²,....,xⁿ = e}, is a subgroup of whatever group x is originally in, say G, then |<x>| divides |G|. but as you can see, |<x>| is the order of x, |x|.

now if a subgroup has order 2, it consists of two elements, one of which is the identity: {e,a}.

a² = a implies that a²*a^-1 = a*a^-1 = e.

but a²*a^-1 = a^2-1 = a¹ = a. so if a² = a, a = e.

but in a two-element group {e,a}, a is DIFFERENT than e. therefore, a², which is certainly in the subgroup {e,a} by closure, can't be a, so it has to be e. this means that a is of order 2.

which means that a² = e.

in the group Z₂₉* under multiplication mod 29, the identity is 1. so if x is an element of order two, we have x² = 1. thus in the FIELD (Z₂₉, +, *), we have that :

x² = 1
x² - 1 = 1 - 1
x² - 1 = 0.

in a field, it is a theorem that a polynomial of degree n cannot have more that n solutions. since x² - 1 = (x - 1)(x + 1), and Z₂₉ is indeed a field, one of x - 1 or x + 1 must be 0, for their product to be 0. so our two solutions are: {1,-1} = {1,28}.

this is the desired subgroup H∩K you were looking for, in the first place.

Quote:

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Originally Posted by Deveno

if a group is finite...

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Whoops, I failed to mention that bit.