Recent content by mathusers

(1): Find all irreducible polynomials of the form x^2 + ax +b , where a,b belong to the field \mathbb{F}_3 with 3 elements. Show explicitly that \mathbb{F}_3(x)/(x^2 + x + 2) is a field by computing its multiplicative monoid. Identify [\mathbb{F}_3(x)/(x^2 + x + 2)]* as an abstract group...

is this correct? [SIZE="2"](iv) Burnside's Lemma states that the center is non-trivial. (using part 2) (i.e. Z(G) \neq {1}), forming the factor group we have a cyclic group. Thus the original group needs to be abelian, including p^2.

sorry i meant (iv), any suggestions there? i guess it follows on from the previous question...

thanks, didnt really expect it to be that easy. i thought there must have been something else to it, but i guess that's maths for us lol how about question (v)? any hints on that please?

that is merely restating the question.. im fairly new to algebra topics.. I've searched here and there but its still a little hazy.. why can't G/Z(G) be cyclic if G is non abelian? are there any hints or explanations behind the dynamics?? ill attempt the proof myself if i get a clearer...

(iii) ok here is the contrapositive. we let H=\text{Z}(G). now if G/H is cyclic then there is aH which generates the group G/H. We Let x,y\in G. Also, Note xH,yH\in G/H thus xH=a^nH and yH=a^mH. This means x = a^n z_1 and y=a^mz_2 where z_1,z_2\in H. But then xy = a^n z_1 a^mz_2 =...

(iii) but wouldn't proving the contrapositive be the OPPOSITE of what is required for the question?

Hi next one, bit confused with this problem: any hints on any of the parts would be greatly appreciated. QUESTION: --------------------------------------- let G be a group of order 8 and suppose that y \epsilon G has ord(y)=4. Put H = [1,y,y^2,y^3] and let x \epsilon G-H (i) show that H...

Hey there, i have a question on the center of a group, regarding group theory. QUESTION: -------------------------------------- The centre Z(G) of a group G is defined by Z(G) = g \epsilon G: \forall x \epsilon G, xg = gx (i) Show that Z(G) is normal subgroup of G (ii) By considering...

[SIZE="3"]ok let me try part (ii) G = D_8 my working: ------------------------------------------------------------------------ Firstly D_8 = [1,x,x^2,x^3,y,xy,x^2y,x^3y] x^4 = 1, y^2 = 1, yx = x^3y If we calculate the orbits then we have: orbit of <1> = {1} <x>: (1)x(1^{-1}) = x...

just a question here though, when it says g_1H = H = Hg_1, is it referring to g_1 as the identity element of the group? if so, can you please explain why this proves that H is normal in G? i understand "how" to do the question but i don't understand "why" it works... thanks :)

ok here is what i have managed to get so far By Sylow's 1st theorem, G has a subgroup of order 11. let n_p be the number of sylow p-subgroups. then n_{11} = 1(mod 11) and n_{11} divides 2^2 (=4) so n_{11}=1. Therefore, it must be a normal subgroup (since it has no distinct conjugates).How can...

[G/H]=2 means that H has two left and right cosets in G. Assume g_1 \in H, then it is trivial that g_1H = H = Hg_1. Now assume g_2 \notin H, this means that g_2H \neq H \neq Hg_2. But since there are only 2 cosets and both of them are not in H then it means they are the same so g_2H = Hg_2 is...

Hi last one here. Any hints on this is appreciated too :) Let G be a group of order 44. Show using Sylow's counting that G has a normal subgroup of order 11. Use the results to classify all groups of order 44.

Hi next one? Any ideas here? Let G be a group and H \subset G a subgroup such that |G/H| = 2. Show that H is normal in G. thnx :)