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Any normal subgroup of order 2 is contained in the centre of the group.(T/F)?
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Any normal subgroup of order 2 is contained in the centre of the group...
Statement:
Any normal subgroup of order 2 is contained in the centre of the group. (T/F)

Answer:
Introduction:
To determine whether the statement is true or false, we need to understand the concepts of normal subgroup, order of a subgroup, and the centre of a group.

Definitions:
1. Normal subgroup: A subgroup H of a group G is called a normal subgroup if for every g ∈ G, gH = Hg, where gH = {gh : h ∈ H} and Hg = {hg : h ∈ H}.
2. Order of a subgroup: The order of a subgroup H, denoted by |H|, is the number of elements in the subgroup.
3. Centre of a group: The centre of a group G, denoted by Z(G), is the set of elements that commute with every element in the group, i.e., Z(G) = {g ∈ G : gh = hg, ∀h ∈ G}.

Proof:
To prove the given statement, we need to show that any normal subgroup of order 2 is contained in the centre of the group.

Case 1: Normal subgroup of order 2:
Let H be a normal subgroup of order 2. This means |H| = 2 and H = {e, a}, where e is the identity element and a is a non-identity element.

Case 2: Commutation with every element:
Let's consider an arbitrary element g ∈ G. We need to show that gh = hg for every h ∈ H.

Subcase 2.1: g ∈ H:
If g is an element of H, then gh = hg for every h ∈ H. This is because H is a subgroup, and by the definition of a subgroup, every element in H commutes with each other.

Subcase 2.2: g ∉ H:
If g is not an element of H, then we need to show that gh = hg for every h ∈ H.

Subcase 2.2.1: gh = hg for h = e:
Since H is a subgroup, e is an element of H. Therefore, we have ge = eg, which implies g = g. This is trivially true.

Subcase 2.2.2: gh = hg for h = a:
Since H is a normal subgroup, ga = ag for every a ∈ H. Therefore, we have ga = ag, which implies g^{-1}ga = g^{-1}ag. Simplifying this expression, we get a = a. This is also trivially true.

Conclusion:
In both subcases 2.2.1 and 2.2.2, we have shown that gh = hg for every h ∈ H. Therefore, the normal subgroup H of order 2 is contained in the centre of the group G.

Final Remarks:
Hence, the given statement is true. Any normal subgroup of order 2 is indeed contained in the centre of the group.
Community Answer
Any normal subgroup of order 2 is contained in the centre of the group...
Statement: Any normal subgroup of order 2 is contained in the centre of the group.

Explanation:
In order to determine whether the statement is true or false, we need to understand the definitions of a normal subgroup, the centre of a group, and the order of a subgroup.

Definition: A subgroup H of a group G is said to be normal if gH = Hg for all g ∈ G, where gH = {gh | h ∈ H} and Hg = {hg | h ∈ H}. In other words, a normal subgroup is one that is invariant under conjugation by elements of the group.

Definition: The centre of a group G, denoted by Z(G), is the set of elements that commute with every element of G. Formally, Z(G) = {g ∈ G | gh = hg for all h ∈ G}.

Definition: The order of a subgroup H of a group G is the number of elements in H, denoted by |H|.

Proof:
Let N be a normal subgroup of order 2 in a group G. We need to show that N ⊆ Z(G), i.e., every element of N commutes with every element of G.

Let x be an arbitrary element of N. Since N is a subgroup of G, x ∈ G. We need to show that x commutes with every element g ∈ G.

Consider an arbitrary element g ∈ G. We need to show that gx = xg.

Since N is a normal subgroup, we know that gN = Ng for all g ∈ G.

Since the order of N is 2, it contains the identity element e and one more element, which we will denote by x.

Therefore, gN = Nx for all g ∈ G.

Now, consider the product gx. Since x ∈ N, we can write x = nx' for some n ∈ N and x' ∈ N.

Thus, gx = g(nx') = (gn)x'.

Since gN = Nx, we can write gn = n'x for some n' ∈ N and x ∈ N.

Therefore, gx = (gn)x' = (n'x)x' = n'(xx') = n'e = ne = n ∈ N.

Similarly, we can show that xg ∈ N.

Thus, gx = xg for all g ∈ G, which implies that x ∈ Z(G).

Since x was an arbitrary element of N, this holds for all elements of N.

Therefore, N ⊆ Z(G).

Conclusion:
From the above proof, we can conclude that any normal subgroup of order 2 is indeed contained in the centre of the group. Hence, the statement is true.
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