Question

If H and K are arbitrary subgroups of G. Prove that HK
is a subgroup of G if and only if HK=KH.

Answer #1

Let H and K be subgroups of a group G so that HK is also a
subgroup. Show that HK = KH.

Let H and K be subgroups of G. Prove that H ∪ K is a subgroup of
G iff H ⊆ K or K ⊆ H.

(a) Prove or disprove: if H and K are subgroups of G, then H ∩ K
is a subgroup of G.
(b) Prove or disprove: if H is an abelian subgroup of G, then G
is abelian

f H and K are subgroups of a group G, let (H,K) be the subgroup
of G generated by the elements {hkh−1k−1∣h∈H, k∈K}.
Show that :
H◃G if and only if (H,G)<H

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,...

(a) Prove or disprove: Let H and K be two normal subgroups of a
group G. Then the subgroup H ∩ K is normal in G. (b) Prove or
disprove: D4 is normal in S4.

Suppose that H is a normal subgroup of G and K is any subgroup
of G. Define HK = {hk : h ? H, k ? K}.
(a) Show that HK is a subgroup of G.
(b) Does the conclusion of (a) continue to hold if H is not
normal in G? Justify

(Abstract algebra) Let G be a group and let H and K be subgroups
of G so that H is not contained in K and K is not contained in H.
Prove that H ∪ K is not a subgroup of G.

Suppose that G is a group with subgroups K ≤
H ≤ G. Suppose that K is normal in
G. Let G act on G/H, the set of
left cosets of H, by left multiplication. Prove that if k
∈ K, then left multiplication of G/H by
k is the identity permutation on G/H.

Prove that if A is a subgroup of G and B is a subgroup of H,
then the direct product A × B is a subgroup of G × H.
Show all steps. Note that AXB is nonempty since the identity e
is a part of A X B. Remains only to show that A X B is closed under
multiplication and inverses.

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