Question

Let G be an abelian group, let H = {x in G | (x^3) = eg}, where eg is the identity of G. Prove that H is a subgroup of G.

Answer #1

**A subgroup is a subset of the group, which is a group
onto itself under the same operation**. To check whether a
given subset is a subgroup we check using one step subgroup test,
two step subgroup test or the finite subgroup test.

In this question we know,

where .

We know, . So . And .

Now, let .

This implies, .

Consider .

Since, the group is abelian, this is equal to

((Since, by definition of H)).

So, since , we get .

**Therefore, H is a subgroup by one step subgroup
test.**

Let G be an Abelian group and H a subgroup of G. Prove that G/H
is Abelian.

Let G be an Abelian group and let H be a subgroup of G Define K
= { g∈ G | g3 ∈ H }. Prove that K is a subgroup of G
.

Let H be a normal subgroup of G. Assume the quotient group G/H
is abelian. Prove that, for any two elements x, y ∈ G, we have x^
(-1) y ^(-1)xy ∈ H

: (a) Let p be a prime, and let G be a finite Abelian group.
Show that Gp = {x ∈ G | |x| is a power of p} is a subgroup of G.
(For the identity, remember that 1 = p 0 is a power of p.) (b) Let
p1, . . . , pn be pair-wise distinct primes, and let G be an
Abelian group. Show that Gp1 , . . . , Gpn form direct sum in...

Suppose that G is a group and H={x|xg=gx for all g∈G}.
a.) Prove that H is a subgroup of G.
b.) Prove that H is abelian.

Let H <| G. If H is abelian and G/H is also abelian, prove or
disprove that G is abelian.

Let G be a group and suppose H = {g5 : g ∈ G} is a
subgroup of G.
(a) Prove that H is normal subgroup of G.
(b) Prove that every element in G/H has order at most 5.

Let G be a finite group and H be a subgroup of G. Prove that if
H is
only subgroup of G of size |H|, then H is normal in G.

Let G be a ﬁnite Abelian group and let n be a positive divisor
of|G|. Show that G has a subgroup of order n.

Let G be a group with subgroups H and K.
(a) Prove that H ∩ K must be a subgroup of G.
(b) Give an example to show that H ∪ K is not necessarily a
subgroup of G.
Note: Your answer to part (a) should be a general proof that the
set H ∩ K is closed under the operation of G, includes the identity
element of G, and contains the inverse in G of each of its
elements,...

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