Give examples of A group G of order 6 that has three 2Sylow

Give examples of: A group G of order 6 that has three 2-Sylow subgroups. A group G that has four 3-Sylow subgroups. Let G be a group of order 35. Prove that G is cyclic. List all Abelian groups, up to isomorphism, of order 72. 4. A group G has order 506 = 2-11-23. Prove that G is not simple. 5. Let Q = {plusminus 1, plusminus i, plusminus j, plusminus k} be the quaternion group. Prove that all subgroups of Q are normal subgroups. Describe all groups, up to isomorphism, of order 70. Let G have order 60. Prove that the center of G cannot have order exactly equal to 4.

Solution

No-5

Let us assume the quaternoin group Q₈={1,-1,i,-i,j,-j,k,-k}

We know that the trivial subgroups satisfies the normal condition trivially.

1 is any subgroup.

{1,-1} form subgroup of order 2

let us choose next three element i , j, or k. we have to include its negative so the three subgroups.

So the subgroups of order 4 are {1,-1,i,-i}, {1,-1,j,-j},{1,-1,k,-k} these are the only possible subgoups of oder 4 for example iff we add j to the first one , then we have to add ij=k and also the negatives which are their inverses.

The next possible subgroup order is 2, so this has to be the subgoup {1,-1}.

We have to show this group is normal.

Now g 1 g^-1 =1.

Next g (-1) g^-1 is either 1 or -1

Also -1 commutes with 1 and -1 .

so we are left considering g (-1) g^-1 where g=i, j , k or -i, -j,, -k

Now i (-1) i^-1=i (-1) (-i) = i²=-1

Similarly j(-1) j^-1 =k (-1) k^-1 =-1

And i (-1) (-i)^-1=(-i)(-1) i =(-i) (-i) = i² =-1

Similarly (-j) (-1) (-j)^-1=(-k)(-1)(-k)^-1=-1

Hence for all 8 in Q_{8 ,} g {1,-1} g^-1 ={1,-1}

So {1,-1} is normal also the subgroup {1} is normal in Q_8.

Therefore Q₈ are normal.