2.2 Groups of order p^2 where p is prime Let p be a prime...

2.2 Groups of order p^2 where p is prime Let p be a prime number and let G be a group with |G| = p^2. The task is to show that G is either cyclic or isomorphic to Zp !! Zp. So, suppose that G is not cyclic. a) Let a ? e be some element of G and A = ?a?. What's the order of A? b) Consider the cosets of A: G/A = {A, g2A, ..., gnA}. What's the index of A in G; that is, the value of n? c) Let ? : G ? SG/A (= the permutation group on the cosets of A) be defined by ?(x)(gkA) = xgkA. Note that ?(x) is a permutation on G/A. First, show that ? is a homomorphism. Now, prove that the only two possibilities for the kernel of ? are 1) ker ? = A or 2) ker ? = {e}. But, case 2) implies that ? is injective. Consider the order of the image ?(G) to show that this can't happen. Now conclude that A is a normal subgroup of G. d) Let b ? G - A. What's the order of b? Since A is normal, bab^-1 = ak for some k, 1 ? k < p. Comparing ?b? to ?a?, establish a^-1ba = bm for some m, 1 ? m < p. e) Derive b^m-1 = a^k-1 and deduce b^m-1 = a^k-1 = e and ab = ba. f) By considering all of the products a^ib^j show that G ? Zp !! Zp. Describe the isomorphism explicitly.

Transcript

00:01 In this question we have been given that p be a prime number and we have been given a group with the order p square.

00:08 The task is to show that g is either cyclic or isomorphic to.

00:13 So g is cyclic or g is isomorphic to zp cross zp.

00:23 Okay, so this is what we have to show.

00:26 So what we will do, we will assume that g is not cyclic.

00:29 This is also given so we are considering g is not cyclic so this thing we have to show that means g isomorphic to zp cross zp we have to so so first of all let's go step by step here what we are going to do so order of if i have to find out what is order of the center of this group okay so it is the number of elements in the center correct so it can be either either p or p square correct also we know that the center of the group is a normal subgroup of g correct so with this what we can see and order of g over z zg that is the center of the group it is one or p because it is the only divisor of p okay so what does that mean so in this case my group g mod zg will become cyclic and when this is cyclic so this implies that g is going to be an abelian group correct so we can say that g is isomorphic to zp square or this is same as the cartesian product of zp and zp correct but we know that g is not cyclic so this will implies that g is going to be isomorphic to zp cross zp only correct now let's see the next part so this is how we saw that g isomorphic to zp cross zp now let's move to the next part in the next part we have been given that if a not equal to e is an element of g okay so this is an element of g and we have been given a set which is generated by this element of g we need to find out what is the order of a correct so order of a will be what so order of a will be same as p only correct since the divisor of p square if we see they are going to be one p and p square if order of a is equal to 1 so this implies a equals to e which is contradiction because a is not equal to e and if order of a is equal to p square so this will implies that g will be cyclic g is cyclic again this is not possible because we already assume that g is not cyclic okay now let's move to the next part in the next part what we have been asked we have been given a cosette a okay so we are considering this as a cosette of a which is consists of the element a g to a and so on up till g n times of a correct what is the index of a in g okay that means we have to basically find out the value of n so let us see how we are going to do this so what we are going to do here so, index of a in g is nothing but i will say it will be order of g divided by order of a.

04:13 So what is order of g? it is p squared divided by order of a, which is p.

04:18 So it comes out to be equal to p.

04:20 So that means value of n comes out to be equal to p.

04:25 And here the order was p.

04:28 Now let's move to the next question in the next question that is part c we are defining a permutation group that is phi defined from g to s g over a okay this is a permutation group on the cosets of a and it is defined as is defined as phi x of gk a is equal to x times gk a okay we have been given that phi of x is a permutation on this first so that phi is homomorphism and then we have to show that the kernel of this phi is equals to a or this is equal to just identity element okay so let us see how we are going to do this so to show that first it is group homomorphism let's see what we can do so first of all we know that we have to show this phi of xy is equals to so this is what we need to show here so five of xy is equals to phi x times phi y so what i will do i will consider one element gk a which belongs to this go set g over a so phi of xy of gk a okay so this can be written as x y xxxxa and this is nothing but x times and this is if you prove this you will get phi y gka correct this is what i will i will get here this can be written as phi of x times phi of y gka which is same as phi x phi y times gka okay and this is same as phi x phi y times gka okay and this is same as phi x gca times phi y gk a.

06:43 This is nothing but phi x gka and phi y gk.

06:49 This is what we needed to prove only.

06:51 Okay.

06:52 So this is what we have shown.

06:54 So from this i can say that phi is a group homomorphism.

07:04 Correct...

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