Which of the following statements are true? Give reasons for your answers.
i) If a group GG is isomorphic to one of its proper subgroups, then G=ZG=\mathbb{Z}.
Answer:
To address this statement, we need to understand the concepts of group isomorphism and proper subgroups, and then apply these concepts to the specific case of the group GG and the integers Z\mathbb{Z}.
Statement: "If a group GG is isomorphic to one of its proper subgroups, then G=ZG = \mathbb{Z}."
A group isomorphism is a bijective group homomorphism between two groups. If there exists an isomorphism between two groups, they are said to be isomorphic, meaning they have the same group structure, but not necessarily the same elements.
A proper subgroup of a group GG is a subgroup HH of GG such that H!=GH \neq G and H!={e}H \neq \{e\}, where ee is the identity element of GG.
Now, let’s analyze the statement:
If GG is isomorphic to one of its proper subgroups: This means there exists a proper subgroup H sub GH \subset G such that there is an isomorphism f:G rarr Hf: G \to H. Since ff is bijective, every element in GG corresponds to a unique element in HH, and vice versa.
Then G=ZG = \mathbb{Z}: This part of the statement claims that the only group for which the above condition can be true is the group of integers under addition, Z\mathbb{Z}.
To justify or refute this statement, we need to consider whether there are any groups other than Z\mathbb{Z} that can be isomorphic to one of their proper subgroups.
Counterexample:
Consider the group Q\mathbb{Q}, the group of rational numbers under addition. Let’s take a proper subgroup of Q\mathbb{Q}, say H={(m)/(2^(n)):m inZ,n inN}H = \{ \frac{m}{2^n} : m \in \mathbb{Z}, n \in \mathbb{N} \}, which is the group of dyadic rationals.
We can define an isomorphism f:Qrarr Hf: \mathbb{Q} \to H by f(q)=2qf(q) = 2q. This function is bijective and preserves the group operation (addition in this case). Therefore, Q\mathbb{Q} is isomorphic to its proper subgroup HH, but Q!=Z\mathbb{Q} \neq \mathbb{Z}.
This counterexample shows that the statement "If a group GG is isomorphic to one of its proper subgroups, then G=ZG = \mathbb{Z}" is false. There exist other groups, like Q\mathbb{Q}, that are also isomorphic to one of their proper subgroups.
ii) If xx and yy are elements of a non-abelian group ( G,**G, * ) such that x**y=y**xx * y=y * x, then x=e\mathrm{x}=\mathrm{e} or y=e\mathrm{y}=\mathrm{e}, where e\mathrm{e} is the identity of G\mathrm{G} with respect to *.
Answer:
The statement to be analyzed is: "If xx and yy are elements of a non-abelian group (G,**)(G, *) such that x**y=y**xx * y = y * x, then x=ex = e or y=ey = e, where ee is the identity of GG with respect to ***."
To evaluate this statement, we need to understand the properties of non-abelian groups and the implications of the given condition x**y=y**xx * y = y * x.
Non-Abelian Group: A group GG is non-abelian if there exist elements in GG such that the group operation is not commutative, i.e., there exist a,b in Ga, b \in G such that a**b!=b**aa * b \neq b * a.
Identity Element: The identity element ee in a group GG is the element that satisfies e**a=a**e=ae * a = a * e = a for all a in Ga \in G.
Now, let’s analyze the statement:
Given: x**y=y**xx * y = y * x for some x,y in Gx, y \in G, a non-abelian group.
To Prove: x=ex = e or y=ey = e.
Counterexample:
To refute the statement, we need to find a non-abelian group GG and elements x,y in Gx, y \in G such that x**y=y**xx * y = y * x but neither xx nor yy is the identity element.
Consider the group of 2xx22 \times 2 invertible matrices under matrix multiplication, which is a non-abelian group. Let’s choose two specific matrices:
These matrices represent 90^(@)90^\circ and -90^(@)-90^\circ rotations in the plane, respectively. It’s easy to verify that x**y=y**xx * y = y * x, as both products represent a 180^(@)180^\circ rotation (or a reflection). However, neither xx nor yy is the identity matrix, which in this case would be:
Thus, we have found a counterexample where x**y=y**xx * y = y * x in a non-abelian group, but neither xx nor yy is the identity element. This disproves the statement, showing that it is false. There can be elements in a non-abelian group that commute with each other without being the identity element.
iii) There exists a unique non-abelian group of prime order.
Answer:
The statement to evaluate is: "There exists a unique non-abelian group of prime order."
To analyze this statement, we need to understand the concepts of group order, abelian groups, and non-abelian groups, particularly in the context of groups of prime order.
Group Order: The order of a group GG is the number of elements in GG. It is denoted as |G||G|.
Abelian Group: A group GG is abelian if for every pair of elements a,ba, b in GG, the group operation is commutative; that is, a**b=b**aa * b = b * a.
Non-Abelian Group: A group is non-abelian if it is not abelian, meaning there exists at least one pair of elements a,ba, b in GG such that a**b!=b**aa * b \neq b * a.
Prime Order: A group of prime order has a number of elements equal to a prime number.
Now, let’s analyze the statement:
Existence of a Non-Abelian Group of Prime Order: The key property of prime numbers is that they have exactly two distinct positive divisors: 1 and themselves. For a group of prime order pp, this means there are pp elements in the group.
Uniqueness and Structure of Groups of Prime Order: According to Lagrange’s Theorem in group theory, the order of every subgroup of a group GG must divide the order of GG. For a group of prime order pp, the only divisors are 1 and pp itself. This implies that the only subgroups of such a group are the trivial group (with just the identity element) and the group itself.
Implication for Abelian Property: In a group of prime order, every element except the identity must generate the entire group (since any subgroup must have order 1 or pp). This means that every non-identity element is a generator of the group. In such a scenario, the group operation must be commutative, as there is only one group structure possible under which all elements (except the identity) are generators. This group structure is essentially cyclic, and all cyclic groups are abelian.
Therefore, any group of prime order must be abelian. This directly contradicts the possibility of a non-abelian group of prime order. Hence, the statement "There exists a unique non-abelian group of prime order" is false. There cannot exist any non-abelian group of prime order, let alone a unique one.
iv) If (a,b)inAxxA(\mathrm{a}, \mathrm{b}) \in \mathrm{A} \times \mathrm{A}, where A\mathrm{A} is a group, then o((a,b))=o(a)o(b)\mathrm{o}((\mathrm{a}, \mathrm{b}))=\mathrm{o}(\mathrm{a}) \mathrm{o}(\mathrm{b}).
Answer:
The statement to evaluate is: "If (a,b)in A xx A(a, b) \in A \times A, where AA is a group, then o((a,b))=o(a)o(b)o((a, b)) = o(a) o(b)."
Here, A xx AA \times A denotes the direct product of the group AA with itself, and o(x)o(x) denotes the order of the element xx in its respective group. The order of an element gg in a group is the smallest positive integer nn such that g^(n)=eg^n = e, where ee is the identity element of the group.
To analyze this statement, we need to understand the order of an element in the direct product of groups and how it relates to the orders of its constituent elements.
Direct Product of Groups: In the direct product A xx AA \times A, the elements are ordered pairs (a,b)(a, b), where a,b in Aa, b \in A. The group operation in A xx AA \times A is defined component-wise. That is, if (a_(1),b_(1)),(a_(2),b_(2))in A xx A(a_1, b_1), (a_2, b_2) \in A \times A, then their product is (a_(1)**a_(2),b_(1)**b_(2))(a_1 * a_2, b_1 * b_2), where *** is the group operation in AA.
Order of an Element in A xx AA \times A: The order of an element (a,b)(a, b) in A xx AA \times A is the smallest positive integer nn such that (a,b)^(n)=(e,e)(a, b)^n = (e, e), where ee is the identity in AA. This means (a^(n),b^(n))=(e,e)(a^n, b^n) = (e, e), which implies a^(n)=ea^n = e and b^(n)=eb^n = e.
Now, let’s analyze the statement:
Given: (a,b)in A xx A(a, b) \in A \times A, where AA is a group.
To Prove: o((a,b))=o(a)o(b)o((a, b)) = o(a) o(b).
Counterexample:
Consider the case where o(a)o(a) and o(b)o(b) are not relatively prime. Let AA be a group, and choose a,b in Aa, b \in A such that o(a)=2o(a) = 2 and o(b)=4o(b) = 4. This means a^(2)=ea^2 = e and b^(4)=eb^4 = e.
In A xx AA \times A, consider the element (a,b)(a, b). We need to find the smallest positive integer nn such that (a,b)^(n)=(e,e)(a, b)^n = (e, e).
For n=2n = 2, (a,b)^(2)=(a^(2),b^(2))=(e,b^(2))(a, b)^2 = (a^2, b^2) = (e, b^2). Since b^(2)!=eb^2 \neq e, n=2n = 2 is not the order of (a,b)(a, b).
For n=4n = 4, (a,b)^(4)=(a^(4),b^(4))=(e,e)(a, b)^4 = (a^4, b^4) = (e, e). Here, n=4n = 4 is the smallest positive integer satisfying the condition.
So, o((a,b))=4o((a, b)) = 4. However, o(a)o(b)=2xx4=8o(a) o(b) = 2 \times 4 = 8.
This counterexample shows that o((a,b))o((a, b)) is not necessarily equal to o(a)o(b)o(a) o(b). The statement is false. The order of an element in the direct product of groups is not always the product of the orders of its constituent elements, especially when the orders of these elements are not relatively prime.
v) If HH and KK are normal subgroups of a group GG, then hk=kh AA h in H,k in Kh k=k h \forall h \in H, k \in K.
Answer:
The statement to evaluate is: "If HH and KK are normal subgroups of a group GG, then hk=khhk = kh for all h in H,k in Kh \in H, k \in K."
To analyze this statement, we need to understand the concept of normal subgroups and their properties in the context of group theory.
Normal Subgroup: A subgroup NN of a group GG is called a normal subgroup if it is invariant under conjugation by elements of GG. This means for every n in Nn \in N and g in Gg \in G, the element gng^(-1)gng^{-1} is also in NN.
Given that HH and KK are normal subgroups of GG, we need to determine whether the commutativity hk=khhk = kh holds for all h in Hh \in H and k in Kk \in K.
Analysis:
Given: HH and KK are normal subgroups of GG.
To Prove: hk=khhk = kh for all h in Hh \in H and k in Kk \in K.
Since HH and KK are normal in GG, for any h in Hh \in H and k in Kk \in K, and for any g in Gg \in G, we have ghg^(-1)in Hghg^{-1} \in H and gkg^(-1)in Kgkg^{-1} \in K.
Now, consider the product khkh. We want to show that this is equal to hkhk. Since HH is normal, k^(-1)hk in Hk^{-1}hk \in H. Let’s denote k^(-1)hkk^{-1}hk by h^(‘)h’. So, h^(‘)=k^(-1)hkh’ = k^{-1}hk, which implies hk=kh^(‘)hk = kh’.
Similarly, since KK is normal, h^(-1)kh in Kh^{-1}kh \in K. Let’s denote h^(-1)khh^{-1}kh by k^(‘)k’. So, k^(‘)=h^(-1)khk’ = h^{-1}kh, which implies kh=hk^(‘)kh = hk’.
Now, we need to show that h^(‘)=hh’ = h and k^(‘)=kk’ = k. Since h^(‘)in Hh’ \in H and h in Hh \in H, and HH is a subgroup, hh^(‘(-1))in Hhh’^{-1} \in H. But hh^(‘(-1))=hk^(-1)h^(-1)khh’^{-1} = hk^{-1}h^{-1}k. Since KK is normal, hk^(-1)h^(-1)in Khk^{-1}h^{-1} \in K, and thus hk^(-1)h^(-1)k in Khk^{-1}h^{-1}k \in K. But this is in HH as well, so it must be the identity element ee, as H nn K={e}H \cap K = \{e\} for normal subgroups HH and KK. Therefore, hh^(‘(-1))=ehh’^{-1} = e, which implies h^(‘)=hh’ = h.
A similar argument shows that k^(‘)=kk’ = k. Therefore, hk=khhk = kh.
This proves that the statement "If HH and KK are normal subgroups of a group GG, then hk=khhk = kh for all h in H,k in Kh \in H, k \in K" is true. The normality of the subgroups HH and KK in GG ensures the commutativity of their elements.