Prove, by contradiction, that A4 has no subgroup of order 6.
To prove, by contradiction, that the alternating group A4 has no subgroup of order 6, let's assume the opposite, i.e., suppose A4 does have a subgroup of order 6. We will derive a contradiction from this assumption.
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What is the order of
If U (R) denotes the group of units of a ring R, show that U (R1 × R2) = U(R1) × U(R2) for rings R1 and R2.
Use FTH to determine all homomorphic images of D8, upto isomorphism.
Let R be a ring, I an ideal of R, J an ideal of I. Show that if J has a unity, then J is an ideal of R. Also give an example to show that if J does not satisfy this condition it need not be an ideal of R.
Prove that every ideal I of a ring R is the kernel of a ring homomorphism of R.
a) Check whether H = {
x= 1 or x isirrational } and K = { x
x
} are subgroups of
b) Let U() { } n = m ∈ |(m, n) =1,m≤ n . Then U(n) is a group with respect to multiplication modulo n. Find the orders of
for each m ∈ U(10).
c) Find Z(D2n), where D2n is the dihedral group with 2n elements,
i) when n is an odd integer;
ii) when n is an even integer.
Check whether the subgroup of reflections and subgroup of rotations in D2n is normal in D2n or not. (Note that D2n is the group of symmetries of an n-gon.)