Group Theory Practice Questions

We are given two operators $\diamond$ and $\square$ defined for positive integers:
$a \diamond b = a + 2b$
$a \square b = ab$

Let's test each statement:
(a) Operator $\diamond$ obeys the associative law:
Associative law for $\diamond$: $(a \diamond b) \diamond c = a \diamond (b \diamond c)$
LHS: $(a \diamond b) \diamond c = (a+2b) \diamond c = (a+2b) + 2c = a+2b+2c$.
RHS: $a \diamond (b \diamond c) = a \diamond (b+2c) = a + 2(b+2c) = a + 2b + 4c$.
Since $a+2b+2c \neq a+2b+4c$ (unless $2c = 4c \Rightarrow c=0$, but values are positive integers), $\diamond$ is NOT associative. So (a) is FALSE.

(b) Operator $\square$ obeys the associative law:
Associative law for $\square$: $(a \square b) \square c = a \square (b \square c)$
LHS: $(a \square b) \square c = (ab) \square c = (ab)c = abc$.
RHS: $a \square (b \square c) = a \square (bc) = a(bc) = abc$.
Since $abc = abc$, $\square$ IS associative. So (b) is TRUE.

(c) Operator $\diamond$ over the operator $\square$ obeys the distributive law:
Distributive law: $a \diamond (b \square c) = (a \diamond b) \square (a \diamond c)$
LHS: $a \diamond (b \square c) = a \diamond (bc) = a + 2(bc) = a+2bc$.
RHS: $(a \diamond b) \square (a \diamond c) = (a+2b) \square (a+2c) = (a+2b)(a+2c) = a^2 + 2ac + 2ab + 4bc$.
Since $a+2bc \neq a^2 + 2ac + 2ab + 4bc$, $\diamond$ does NOT distribute over $\square$. So (c) is FALSE.

(d) Operator $\square$ over the operator $\diamond$ obeys the distributive law:
Distributive law: $a \square (b \diamond c) = (a \square b) \diamond (a \square c)$
LHS: $a \square (b \diamond c) = a \square (b+2c) = a(b+2c) = ab+2ac$.
RHS: $(a \square b) \diamond (a \square c) = (ab) \diamond (ac) = (ab) + 2(ac) = ab+2ac$.
Since $ab+2ac = ab+2ac$, $\square$ DOES distribute over $\diamond$. So (d) is TRUE.

Therefore, statements (b) and (d) are TRUE.

An element $(a, b, c) \in \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_4$ is its own inverse if $2a \equiv 0 \pmod 2$, $2b \equiv 0 \pmod 3$, and $2c \equiv 0 \pmod 4$.
For $a \in \mathbb{Z}_2$, the equation $2a \equiv 0 \pmod 2$ is satisfied by all $a \in \{0, 1\}$, giving 2 choices.
For $b \in \mathbb{Z}_3$, the equation $2b \equiv 0 \pmod 3$ implies $b \equiv 0 \pmod 3$ because $\gcd(2, 3) = 1$, giving 1 choice.
For $c \in \mathbb{Z}_4$, the equation $2c \equiv 0 \pmod 4$ is satisfied by $c \in \{0, 2\}$, giving 2 choices.
The total number of elements that are their own inverses is the product of the number of choices for each component:
$2 \times 1 \times 2 = 4$

Let's analyze each statement for a group G:
(A) "If for all $x, y \in G, (xy)^2 = x^2y^2$, then G is commutative."
Given $(xy)^2 = x^2y^2$.
This expands to $xyxy = xxyy$.
By multiplying by $x^{-1}$ on the left, we get $yxy = xyy$.
By multiplying by $y^{-1}$ on the right, we get $yx = xy$.
This is the definition of a commutative group. So, statement (A) is TRUE.

(B) "If for all $x \in G, x^2 = 1$, then G is commutative. Here, 1 is the identity element of G."
Given $x^2 = 1$ for all $x \in G$. This implies $x = x^{-1}$.
Consider any two elements $x, y \in G$.
We know that $(xy)^2 = 1$ (since $xy$ is also an element of G).
So, $xyxy = 1$.
Since $x=x^{-1}$ and $y=y^{-1}$, we also have $(xy)^{-1} = xy$.
We know that $(xy)^{-1} = y^{-1}x^{-1}$.
So, $xy = y^{-1}x^{-1}$.
Substituting $y^{-1}=y$ and $x^{-1}=x$, we get $xy = yx$.
Thus, G is commutative. So, statement (B) is TRUE.

(C) "If the order of G is 2, then G is commutative."
If the order of G is 2, then G has two elements: the identity element $e$ and one other element, say $a$. So $G = \{e, a\}$.
In any group, $e$ commutes with all elements ($ex = xe = x$).
We need to check if $ae = ea$. This is true.
Thus, a group of order 2 is always commutative (in fact, any group of prime order is cyclic, and all cyclic groups are abelian/commutative). So, statement (C) is TRUE.

(D) "If G is commutative, then a subgroup of G need not be commutative."
If G is commutative, then for any $x, y \in G$, $xy = yx$.
Let H be any subgroup of G. Then for any $x, y \in H$, $x, y$ are also elements of G.
Since $x, y \in G$, we have $xy = yx$.
Therefore, H is also commutative.
So, this statement is FALSE (a subgroup of a commutative group must also be commutative).

The TRUE statements are (A), (B), and (C).

Given $|G|=6$. By Lagrange's theorem, the order of any subgroup $H$ must divide the order of $G$. Thus, $|H|$ can be $1, 2, 3,$ or $6$.
The problem states that $1 < |H| < 6$, which implies that $|H|$ must be $2$ or $3$.
Since both $2$ and $3$ are prime numbers, any group of prime order is cyclic. Therefore, $H$ is always cyclic.
However, $G$ has order $6$, which is not a prime number. Hence, $G$ may or may not be cyclic.

The problem asks for the largest possible size of a subgroup of a group G, given that G has 35 elements, and the subgroup must not be G itself.

1. We are given a group G with $|G| = 35$ elements.
2. Lagrange's Theorem states that for any finite group G, the order (number of elements) of any subgroup H of G must divide the order of G.
3. In this case, the order of G is 35. We need to find the divisors of 35. The divisors of 35 are 1, 5, 7, and 35.
4. A subgroup H can have an order of 1, 5, 7, or 35.
5. The problem specifies that the subgroup must be "other than G itself." This means we exclude the case where the subgroup has an order equal to the order of G, which is 35.
6. The remaining possible orders for a subgroup H (other than G itself) are 1, 5, and 7.
7. The largest among these remaining possible orders is 7.

The largest possible size of a subgroup of G other than G itself is 7.

[CORRECT_OPTION: N/A]

According to Lagrange's theorem, the order (number of elements) of any subgroup of a finite group must be a divisor of the order of the group. The order of the group G is 84.

The divisors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.

A "proper" subgroup is any subgroup of G except G itself. Therefore, the order of a proper subgroup must be a divisor of 84 that is strictly less than 84. The largest such divisor is 42.

Let $e$ be the identity element.
Given $x*x = e$, so $x$ is its own inverse, i.e., $x^{-1} = x$.
Given $y*y = e$, so $y$ is its own inverse, i.e., $y^{-1} = y$.
Given $x*y*x*y = e$. Since $y^{-1}=y$, we can write this as $x*y*x = y^{-1} = y$.
Multiplying by $x$ on the left, we get $x*x*y*x = x*y$, which simplifies to $e*y*x = x*y$, so $y*x = x*y$.
Since $y*x = x*y$, the group is abelian.
The elements of the group are products of $x$ and $y$. The possible distinct elements are $e, x, y, x*y$.
$x*y$ is distinct because if $x*y = e$, then $x=y^{-1}=y$, which would contradict $x*y*x*y=e$ (unless $x=y=e$). If $x*y=x$, then $y=e$, if $x*y=y$, then $x=e$.
Given $x,y$ are elements, we assume they are not necessarily $e$. If $x \neq e$ and $y \neq e$, and $x \neq y$, then the group contains $e, x, y, xy$.
The four elements $e, x, y, xy$ form the group.
[CORRECT_OPTION: 4]

By Lagrange's Theorem, the order (size) of a subgroup must divide the order of the group. Here, the order of group $G$ is $|G| = 15$.

Let $|L|$ be the size of the subgroup $L$. Therefore, $|L|$ must be a divisor of 15. The divisors of 15 are 1, 3, 5, and 15.

We are given that $L \neq G$, which implies that $|L| \neq |G|$. Thus, $|L| \neq 15$. The possible sizes for $L$ are now $\{1, 3, 5\}$.

We are also given that the size of $L$ is at least 4, which means $|L| \ge 4$.

Comparing this condition with the remaining possibilities $\{1, 3, 5\}$, the only value that satisfies $|L| \ge 4$ is 5. Therefore, the size of $L$ must be 5.

[CORRECT_OPTION: 5]

To determine the algebraic structure of (S, ), we check if it satisfies the four axioms of a group. The set is S = {1, $\omega, \omega^2$}, where $\omega$ and $\omega^2$ are complex cube roots of unity, which means $\omega^3 = 1$. The operation is multiplication ().

1. Closure: The product of any two elements in S must also be in S.

   • $\omega \times \omega = \omega^2 \in S$
   • $\omega \times \omega^2 = \omega^3 = 1 \in S$
   • All other products (e.g., $1 \times \omega = \omega$) are also in S. The property holds.

2. Associativity: Multiplication of complex numbers is associative. For any elements a, b, c in S, $(a * b) * c = a * (b * c)$. This property holds.

3. Identity Element: There must be an element 'e' in S such that $a * e = a$ for all 'a' in S. The number 1 is the identity element in this set, as $1 * x = x$ for any x in S.

4. Inverse Element: For each element 'a' in S, there must be an inverse 'a⁻¹' in S such that $a * a⁻¹ = 1$.

   • The inverse of 1 is 1 (since $1 * 1 = 1$).
   • The inverse of $\omega$ is $\omega^2$ (since $\omega * \omega^2 = 1$).
   • The inverse of $\omega^2$ is $\omega$ (since $\omega^2 * \omega = 1$).

Since all four properties are satisfied, the structure (S, *) forms a group. It cannot be a ring, integral domain, or field as these structures require two operations (e.g., addition and multiplication), and only one operation is defined here.

A group $(G, *)$ is defined by four axioms: closure, associativity, existence of an identity element, and existence of an inverse for every element.

1. Associativity is a fundamental axiom.
2. Existence of an identity element is a fundamental axiom.
3. Existence of an inverse for every element is a fundamental axiom.
4. Commutativity ($a * b = b * a$) is not a required property for a general group. If a group satisfies commutativity, it is called an abelian group. Therefore, not all groups are commutative.

The identity element is $a$, as $a * x = x$ and $x * a = x$ for all $x$. The group has order 4.
An element $g$ is a generator if its order is 4.

1. Order of $b$: $b^1 = b$, $b^2 = b * b = a$. Order of $b$ is 2.
2. Order of $c$: $c^1 = c$, $c^2 = c * c = b$, $c^3 = b * c = d$, $c^4 = d * c = a$. Order of $c$ is 4.
3. Order of $d$: $d^1 = d$, $d^2 = d * d = b$, $d^3 = b * d = c$, $d^4 = c * d = a$. Order of $d$ is 4.
   Since $c$ and $d$ are the only elements with order 4, they are the generators of the group.

To find the number of non-isomorphic Abelian groups of order 4, we use the Fundamental Theorem of Finite Abelian Groups.
First, find the prime factorization of the order: $4 = 2^2$.
The number of non-isomorphic Abelian groups of order $p^k$ is equal to the number of partitions of $k$. Here, $p=2$ and $k=2$.
The partitions of 2 are:

1. $2$: This corresponds to the cyclic group $\mathbb{Z}_4$.
2. $1+1$: This corresponds to the direct product $\mathbb{Z}_2 \times \mathbb{Z}_2$.
   These two groups are non-isomorphic (e.g., $\mathbb{Z}_4$ has an element of order 4, while $\mathbb{Z}_2 \times \mathbb{Z}_2$ does not).
   Thus, there are 2 different non-isomorphic Abelian groups of order 4.