Test: Group Theory - 7 - Mathematics MCQ

To find number of Sylow subgroup of G of order 3.
15 = 5 x 3, by a known theorem.
Number of 3-Sylow subgroup = 3K + 1
Now, (3K + 1) must divide 15, (3K + 1) being prime to 3 should divide 5. Thus, is true only for K = 0 thus, 37K+1 = 1.
Therefore, Number of 3-Sylow subgroup = 1.

we have (1 2 3) (1 2) = (1 2) (1 3) (1 2) which is a product of odd number of transpositions.

Indeed, for integers m,n (not both zero),
⟨m,n⟩ = gcd(m,n)Z
In this case, that means that
⟨2,7⟩ = gcd(2,7)Z
= 1Z
= Z

Let G = A₄, H = {(1 2)(3 4),(1 3)(2 4),(1 4)(2 3), e},

K = {e, (1 2) (3 4)} 

Then H is normal in G and K is normal in H. But K is not normal in G

we have o(HK) = 
Implies 
Implies o(H ∩ K) = 3

By the Lagrange’s theorem, the order of subgroup divides the order of group. We claim that all elements of G cannot be of order 2. Suppose it so.

Let a,b ∈ G be two different element of order 2. Let H = < a >, K = < b > be the cyclic groups generated by a and b, then
o(H) = 2, o(K) = 2
Since all the elements of group G are of order 2, it must be abelian.
∴ HK= KH ⇒ HK is subgroup of G And so

By Lagrange’s theorem, o(HK) would divide o(G).Which is not true hence our assumption is not correct.
Again, since G is finite, o(a) I o(G) for all a ∈ G
⇒ there exist at least one element a ∈ G such that  o(a) = 5 or 10.
If o(a) = 5, then H = < a > is subgroup of order 5.

If o(a) = 10, then H = < a2 > is a subgroup of order 5.

Let G be a simple group and f:G→G′ is homomorphism. So the Kernal f is a normal subgroup of G. But only normal subgroups of G are G and {e}.

∴ Either ker f = G or ker f = {e}

⇒⇒ either f is trivial or f is one - one

we have o(AB) = 
Implies 
= 30/10 = 3