Homework 1: Group Theory

Problem Sheet 1 · MTH 301

Instructor: Ajit Bhand · August 10, 2026

Instructions: Give proper justifications for your answers. Merely identifying a familiar structure is not sufficient unless the relevant axioms have been verified. In problems involving proofs, indicate explicitly where associativity, identity, or invertibility is used.

Problems to be submitted: 2, 4, 5, 9, 14.

Submission day: Monday, August 17, 2026, in my office.

Binary operations and Abelian groups

Question 1 · Averaging on the real numbers

Problem

On \(\mathbb{R}\) define

\[ x * y = \frac{x+y}{2}. \]

Determine which of the following properties hold: commutativity, associativity, existence of an identity, and idempotence (\(x*x=x\) for all \(x\)).

Solution

The operation is commutative, since

\[ x * y = \frac{x+y}{2} = \frac{y+x}{2} = y * x. \]

It is idempotent because

\[ x * x = x. \]

It is not associative; for example,

\[ (0 * 0) * 2 = 0 * 2 = 1, \]

whereas

\[ 0 * (0 * 2) = 0 * 1 = \frac{1}{2}. \]

If \(e\) were an identity, then

\[ e * x = \frac{e+x}{2} = x \]

would force \(e = x\) for every \(x\), which is impossible. Thus there is no identity.

Question 2 · Shifted addition on the integers

Problem

Define a binary operation on \(\mathbb{Z}\) by

\[ m * n = m + n + 1. \]

Show that \((\mathbb{Z}, *)\) is an Abelian group.
Find its identity and the inverse of an arbitrary \(m \in \mathbb{Z}\).
Show that
\[ \varphi: (\mathbb{Z}, *) \longrightarrow (\mathbb{Z}, +), \quad \varphi(m) = m+1 \]
is a bijective magma homomorphism.

Solution

We have

\[ (m * n) * r = m + n + r + 2 = m * (n * r), \]

so \(*\) is associative, and it is clearly commutative. The identity \(e\) satisfies

\[ m + e + 1 = m, \]

hence \(e = -1\). The inverse of \(m\) is the integer \(-m-2\).

Moreover,

\[ \varphi(m * n) = m + n + 2 = (m+1) + (n+1) = \varphi(m) + \varphi(n). \]

The map is clearly bijective.

Question 3 · Rational numbers with odd denominator

Problem

Let

\[ S = \left\{\frac{a}{b} \in \mathbb{Q} : a, b \in \mathbb{Z}, b \ne 0, b \text{ odd}\right\}. \]

Prove that \(S\) is an Abelian group under ordinary addition.

Solution

If \(a/b, c/d \in S\) with \(b, d\) odd, then

\[ \frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}, \]

and \(bd\) is odd, so the sum lies in \(S\). Associativity and commutativity are inherited from \(\mathbb{Q}\). The identity is \(0 = 0/1\), and

\[ -\frac{a}{b} = \frac{-a}{b} \in S. \]

Hence \(S\) is an Abelian group.

Question 4 · Symmetric difference of sets

Problem

Let \(X\) be a set. For \(A, B \subseteq X\), define the symmetric difference

\[ A \triangle B = (A \setminus B) \cup (B \setminus A). \]

Prove that \((\mathcal{P}(X), \triangle)\) is an Abelian group.

Solution

Recall that for \(A, B \subseteq X\), the symmetric difference is

\[ A \triangle B = (A \setminus B) \cup (B \setminus A). \]

Equivalently,

\[ A \triangle B = (A \cup B) \setminus (A \cap B). \]

Thus, for \(x \in X\),

\[ x \in A \triangle B \iff x \text{ belongs to exactly one of } A \text{ and } B. \]

We verify the axioms of an abelian group.

Closure. Let \(A, B \in \mathcal{P}(X)\). Then \(A, B \subseteq X\), and hence

\[ A \setminus B \subseteq X \quad \text{and} \quad B \setminus A \subseteq X. \]

Therefore

\[ A \triangle B = (A \setminus B) \cup (B \setminus A) \subseteq X. \]

Thus

\[ A \triangle B \in \mathcal{P}(X), \]

so \(\triangle\) is a binary operation on \(\mathcal{P}(X)\).

Associativity. Let \(A, B, C \in \mathcal{P}(X)\). We prove that

\[ (A \triangle B) \triangle C = A \triangle (B \triangle C). \]

Let \(x \in X\). We determine when \(x\) belongs to \((A \triangle B) \triangle C\). There are four possibilities for the number of sets among \(A, B, C\) which contain \(x\).

Suppose that \(x\) belongs to none of \(A, B, C\). Then
\[ x \notin A \triangle B \quad \text{and} \quad x \notin C, \]
so
\[ x \notin (A \triangle B) \triangle C. \]
Suppose that \(x\) belongs to exactly one of \(A, B, C\). If, for example, \(x \in A\) and \(x \notin B, C\), then
\[ x \in A \triangle B \quad \text{and} \quad x \notin C, \]
and hence
\[ x \in (A \triangle B) \triangle C. \]
The other two cases are analogous.
Suppose that \(x\) belongs to exactly two of \(A, B, C\). For example, suppose
\[ x \in A, \quad x \in B, \quad x \notin C. \]
Since \(x\) belongs to both \(A\) and \(B\),
\[ x \notin A \triangle B. \]
Since also \(x \notin C\),
\[ x \notin (A \triangle B) \triangle C. \]
If instead \(x\) belongs to \(A\) and \(C\) but not \(B\), then
\[ x \in A \triangle B \quad \text{and} \quad x \in C, \]
so again
\[ x \notin (A \triangle B) \triangle C. \]
The remaining case is identical.
Suppose that \(x\) belongs to all three sets \(A, B, C\). Since \(x\) belongs to both \(A\) and \(B\),
\[ x \notin A \triangle B. \]
But \(x \in C\), and therefore
\[ x \in (A \triangle B) \triangle C. \]

Consequently,

\[ x \in (A \triangle B) \triangle C \]

if and only if \(x\) belongs to an odd number of the sets \(A, B, C\); that is, if and only if \(x\) belongs to exactly one of them or to all three.

Applying exactly the same argument, with \(B\) and \(C\) considered first, we obtain

\[ x \in A \triangle (B \triangle C) \]

if and only if \(x\) belongs to an odd number of the sets \(A, B, C\).

Hence, for every \(x \in X\),

\[ x \in (A \triangle B) \triangle C \iff x \in A \triangle (B \triangle C). \]

Therefore

\[ (A \triangle B) \triangle C = A \triangle (B \triangle C). \]

Thus \(\triangle\) is associative.

Identity element. We claim that \(\emptyset\) is the identity element. For every \(A \subseteq X\),

\[ A \triangle \emptyset = (A \setminus \emptyset) \cup (\emptyset \setminus A) = A \cup \emptyset = A. \]

Similarly,

\[ \emptyset \triangle A = A. \]

Hence the identity element is \(\emptyset\).

Inverses. Let \(A \in \mathcal{P}(X)\). Then

\[ A \triangle A = (A \setminus A) \cup (A \setminus A) = \emptyset. \]

Since \(\emptyset\) is the identity element, \(A\) is its own inverse. Thus

\[ A^{-1} = A \]

with respect to the operation \(\triangle\).

Commutativity. For \(A, B \subseteq X\),

\[\begin{aligned} A \triangle B &= (A \setminus B) \cup (B \setminus A) \\ &= (B \setminus A) \cup (A \setminus B) \\ &= B \triangle A. \end{aligned}\]

Hence \(\triangle\) is commutative.

Therefore \(\mathcal{P}(X)\) is a group under symmetric difference, and the group operation is commutative. Hence \((\mathcal{P}(X), \triangle)\) is an abelian group.

Question 5 · The operation x + y + xy

Problem

On \(\mathbb{R}\) define

\[ x * y = x + y + xy. \]

Prove that \(*\) is associative and commutative.
Find the identity element.
Determine exactly which elements of \(\mathbb{R}\) are invertible.
Deduce that \(\mathbb{R} \setminus \{-1\}\) is an Abelian group under \(*\).

Solution

A direct computation gives

\[ (x * y) * z = x + y + z + xy + xz + yz + xyz = x * (y * z). \]

The operation is visibly commutative. The identity is \(0\).

An inverse \(y\) of \(x\) must satisfy

\[ x + y + xy = 0, \]

so, if \(x \ne -1\),

\[ y = -\frac{x}{1+x}. \]

For \(x = -1\) we have

\[ -1 * y = -1 \]

for every \(y\), so \(-1\) is not invertible. Hence the units are precisely \(\mathbb{R} \setminus \{-1\}\), which therefore forms an Abelian group under \(*\).

Function groups and group properties

Question 6 · Pointwise multiplication of functions

Problem

Let \(G\) be a group and let \(X\) be a nonempty set. Let

\[ G^{X} = \{f \mid f : X \rightarrow G\}. \]

Define

\[ (f * g)(x) = f(x)g(x) \quad (x \in X). \]

Prove that \(G^{X}\) is a group under \(*\).
Prove that \(G^{X}\) is Abelian if \(G\) is Abelian.
Prove the converse: if \(G^{X}\) is Abelian, then \(G\) is Abelian.

Solution

Associativity follows pointwise:

\[ ((f * g) * h)(x) = (f(x)g(x))h(x) = f(x)(g(x)h(x)) = (f * (g * h))(x). \]

The identity is the constant function \(x \mapsto e_{G}\), and

\[ f^{-1}(x) = f(x)^{-1}. \]

Thus \(G^{X}\) is a group.

If \(G\) is Abelian, multiplication is pointwise commutative, so \(G^{X}\) is Abelian. Conversely, since \(X\) is nonempty, for any \(a, b \in G\), consider the constant functions \(f_{a}(x) = a\) and \(f_{b}(x) = b\). If \(G^{X}\) is Abelian, then

\[ f_{a}f_{b} = f_{b}f_{a}, \]

hence \(ab = ba\). Thus \(G\) is Abelian.

Question 7 · Affine transformations of the real line

Problem

Let \(A\) be the set of all functions

\[ f_{a,b} : \mathbb{R} \rightarrow \mathbb{R}, \quad f_{a,b}(x) = ax + b, \]

where \(a \in \mathbb{R} \setminus \{0\}\) and \(b \in \mathbb{R}\).

Show that \(A\) is a group under composition.
Find the inverse of \(f_{a,b}\).
Show that this group is not Abelian.

Solution

We have

\[ (f_{a,b} \circ f_{c,d})(x) = a(cx + d) + b = acx + (ad + b), \]

and \(ac \ne 0\), so \(A\) is closed under composition. Associativity is inherited from composition of functions. The identity is \(f_{1,0}\). Solving

\[ y = ax + b \]

for \(x\) gives

\[ f_{a,b}^{-1}(y) = \frac{1}{a}y - \frac{b}{a}, \]

\[ f_{a,b}^{-1} = f_{1/a, -b/a}. \]

The group is not Abelian; for instance, with

\[ f(x) = 2x, \quad g(x) = x + 1, \]

we obtain

\[ (f \circ g)(x) = 2x + 2, \quad (g \circ f)(x) = 2x + 1. \]

Question 8 · Groups whose elements square to the identity

Problem

Let \(G\) be a group in which

\[ x^{2} = e \quad \text{for every } x \in G. \]

Prove that \(G\) is Abelian.

Solution

Since \(x^{2} = e\), every element satisfies \(x^{-1} = x\). Hence for \(a, b \in G\),

\[ ab = (ab)^{-1} = b^{-1}a^{-1} = ba. \]

Therefore \(G\) is Abelian.

Question 9 · A test for an Abelian group

Problem

Let \(G\) be a group. Prove that the following are equivalent:

\[ G \text{ is Abelian} \quad \text{and} \quad (ab)^{2} = a^{2}b^{2} \text{ for all } a, b \in G. \]

Solution

If \(G\) is Abelian, then

\[ (ab)^{2} = abab = aabb = a^{2}b^{2}. \]

Conversely, suppose

\[ abab = aabb \]

for all \(a, b\). Left cancellation by \(a\) gives

\[ bab = abb, \]

and right cancellation by \(b\) gives

\[ ba = ab. \]

Thus \(G\) is Abelian.

Question 10 · The multiplication table of a three-element group

Problem

Suppose \(G = \{e, a, b\}\) is a group with exactly three elements. Without using any theorem about cyclic groups or the order of an element, determine the entire multiplication table of \(G\).

Solution

Since \(e\) is the identity, the first row and column are determined. We cannot have \(a^{2} = a\), for then cancellation in

\[ a^{2} = ae \]

would give \(a = e\). If \(a^{2} = e\), then the row of \(a\) would contain

\[ a, \ e, \ ab. \]

Since each element occurs once in the row, \(ab = b\). Right cancellation in

\[ ab = eb \]

would then give \(a = e\), a contradiction. Therefore

\[ a^{2} = b. \]

Similarly,

\[ b^{2} = a. \]

The remaining products are:

\[ ab = ba = e. \]

Thus

\[ \begin{array}{c|ccc} \cdot & e & a & b \\ \hline e & e & a & b \\ a & a & b & e \\ b & b & e & a \end{array} \]

Inverses and constructed operations

Question 11 · One-sided identity and one-sided inverses

Problem

Let \(S\) be an associative system with an element \(e\) satisfying

\[ ex = x \quad \text{for every } x \in S. \]

Assume also that for every \(x \in S\) there exists \(y \in S\) such that

\[ yx = e. \]

Prove that \(S\) is a group.

Solution

Fix \(x \in S\), and choose \(y\) with \(yx = e\). Choose \(z\) with \(zy = e\). Then

\[ z = ze = z(yx) = (zy)x = ex = x. \]

Hence \(z = x\), so

\[ xy = zy = e. \]

Thus \(y\) is also a right inverse of \(x\).

Now

\[ xe = x(yx) = (xy)x = ex = x. \]

Hence \(e\) is also a right identity. Every element therefore has a two-sided inverse and \(e\) is a two-sided identity. Together with associativity, this shows that \(S\) is a group.

Question 12 · Invertible products in a monoid

Problem

Let \(M\) be a monoid and let \(a, b \in M\). Assume that both \(ab\) and \(ba\) are invertible. Prove that \(a\) and \(b\) are invertible.

Solution

Since \(ab\) is invertible,

\[ a(b(ab)^{-1}) = e, \]

so \(a\) has a right inverse. Since \(ba\) is invertible,

\[ ((ba)^{-1}b)a = e, \]

so \(a\) has a left inverse. In any monoid, a left inverse and a right inverse of the same element coincide: if \(ua = e\) and \(av = e\), then

\[ u = u(av) = (ua)v = v. \]

Hence \(a\) is invertible.

Similarly, \(b\) has a left inverse obtained from \(ab\) and a right inverse obtained from \(ba\), so \(b\) is invertible.

Question 13 · Classifying affine binary operations

Problem

Fix real numbers \(\alpha, \beta, \gamma\). Define

\[ x * y = \alpha x + \beta y + \gamma \quad (x, y \in \mathbb{R}). \]

Classify all triples \((\alpha, \beta, \gamma)\) for which:

\(*\) is associative;
\(*\) is commutative;
\((\mathbb{R}, *)\) is a monoid;
\((\mathbb{R}, *)\) is a group.

Solution

We compute

\[ (x * y) * z = \alpha^{2}x + \alpha\beta y + \alpha\gamma + \beta z + \gamma, \]

whereas

\[ x * (y * z) = \alpha x + \alpha\beta y + \beta^{2}z + \beta\gamma + \gamma. \]

Thus associativity is equivalent to

\[ \alpha^{2} = \alpha, \quad \beta^{2} = \beta, \quad \gamma(\alpha - \beta) = 0. \]

Commutativity is equivalent to

\[ \alpha = \beta. \]

Suppose there is a two-sided identity \(e\). From

\[ e * x = x \]

for every \(x\), we get

\[ \beta = 1, \quad \alpha e + \gamma = 0. \]

From

\[ x * e = x \]

we get

\[ \alpha = 1, \quad \beta e + \gamma = 0. \]

Hence a monoid occurs exactly when

\[ \alpha = \beta = 1, \]

with arbitrary \(\gamma\), and then

\[ e = -\gamma. \]

For these parameters,

\[ x * y = x + y + \gamma. \]

The inverse of \(x\) is

\[ -x - 2\gamma. \]

Hence every such monoid is in fact an Abelian group. Therefore the group classification is again

\[ \alpha = \beta = 1, \quad \gamma \in \mathbb{R}. \]

Question 14 · Parity-twisted addition on the integers

Problem

Define an operation on \(\mathbb{Z}\) by

\[ m * n = m + (-1)^{m}n. \]

Prove that \(*\) is associative.
Find the identity.
Find the inverse of an arbitrary integer \(m\).
Show that the resulting group is non Abelian.

Solution

First note that

\[ (-1)^{m+(-1)^{m}n} = (-1)^{m+n} \]

because \((-1)^{m}n\) has the same parity as \(n\). Hence

\[ (m * n) * r = m + (-1)^{m}n + (-1)^{m+n}r. \]

On the other hand,

\[ m * (n * r) = m + (-1)^{m}(n + (-1)^{n}r) = m + (-1)^{m}n + (-1)^{m+n}r. \]

Thus \(*\) is associative.

The identity is \(0\). To solve

\[ m * n = 0, \]

we need

\[ m + (-1)^{m}n = 0. \]

Thus

\[ n = \begin{cases} -m, & m \text{ even}, \\ m, & m \text{ odd}. \end{cases} \]

The same formula gives a left inverse, so this is the two-sided inverse.

Finally,

\[ 1 * 2 = 1 - 2 = -1, \quad 2 * 1 = 2 + 1 = 3. \]

Hence the group is nonAbelian.

Question 15 · The matrix operation A + B + AB

Problem

On \(M_{n}(\mathbb{R})\) define

\[ A * B = A + B + AB. \]

Prove that \(*\) is associative.
Find the identity element.
Show that \(A\) is invertible with respect to \(*\) if and only if \(I_{n} + A\) is invertible under ordinary matrix multiplication.
When \(A\) is invertible with respect to \(*\), express its \(*\)-inverse in terms of \((I_{n} + A)^{-1}\).
For \(n \ge 2\) show that \(*\) need not be commutative.

Solution

For \(A, B \in M_{n}(\mathbb{R})\) define

\[ A * B = A + B + AB. \]

Since \(A + B + AB \in M_{n}(\mathbb{R})\), the operation is a binary operation on \(M_{n}(\mathbb{R})\).

The key identity is

\[ I_{n} + (A * B) = I_{n} + A + B + AB = (I_{n} + A)(I_{n} + B). \]

We shall use this repeatedly.

We prove that \(*\) is associative. Let \(A, B, C \in M_{n}(\mathbb{R})\). Then
\[\begin{aligned} (A * B) * C &= (A + B + AB) + C + (A + B + AB)C \\ &= A + B + C + AB + AC + BC + ABC. \end{aligned}\]
On the other hand,
\[\begin{aligned} A * (B * C) &= A + (B + C + BC) + A(B + C + BC) \\ &= A + B + C + BC + AB + AC + ABC \\ &= A + B + C + AB + AC + BC + ABC. \end{aligned}\]
Therefore
\[ (A * B) * C = A * (B * C). \]
Hence \(*\) is associative.

Equivalently, using the key identity,
\[\begin{aligned} I_{n} + ((A * B) * C) &= (I_{n} + (A * B))(I_{n} + C) \\ &= (I_{n} + A)(I_{n} + B)(I_{n} + C), \end{aligned}\]
while
\[\begin{aligned} I_{n} + (A * (B * C)) &= (I_{n} + A)(I_{n} + (B * C)) \\ &= (I_{n} + A)(I_{n} + B)(I_{n} + C). \end{aligned}\]
Thus the two matrices are equal.
We claim that the zero matrix \(0\) is the identity element. Indeed, for every \(A \in M_{n}(\mathbb{R})\),
\[ A * 0 = A + 0 + A0 = A \]
and
\[ 0 * A = 0 + A + 0A = A. \]
Therefore the identity element for \(*\) is \(0\).
We show that \(A\) is invertible with respect to \(*\) if and only if \(I_{n} + A\) is invertible with respect to ordinary matrix multiplication.

Suppose first that \(A\) is invertible with respect to \(*\). Then there exists \(B \in M_{n}(\mathbb{R})\) such that
\[ A * B = 0 \quad \text{and} \quad B * A = 0. \]
Using
\[ I_{n} + (A * B) = (I_{n} + A)(I_{n} + B), \]
we obtain
\[ (I_{n} + A)(I_{n} + B) = I_{n}. \]
Similarly,
\[ (I_{n} + B)(I_{n} + A) = I_{n}. \]
Thus \(I_{n} + B\) is a two-sided multiplicative inverse of \(I_{n} + A\). Therefore \(I_{n} + A\) is invertible.

Conversely, suppose that \(I_{n} + A\) is invertible. Define
\[ B = (I_{n} + A)^{-1} - I_{n}. \]
Then
\[ I_{n} + B = (I_{n} + A)^{-1}. \]
Hence
\[\begin{aligned} I_{n} + (A * B) &= (I_{n} + A)(I_{n} + B) \\ &= (I_{n} + A)(I_{n} + A)^{-1} \\ &= I_{n}. \end{aligned}\]
Therefore
\[ A * B = 0. \]
Similarly,
\[\begin{aligned} I_{n} + (B * A) &= (I_{n} + B)(I_{n} + A) \\ &= (I_{n} + A)^{-1}(I_{n} + A) \\ &= I_{n}, \end{aligned}\]
so
\[ B * A = 0. \]
Thus \(B\) is a \(*\)-inverse of \(A\).

Consequently, \(A\) is invertible with respect to \(*\) \(\iff\) \(I_{n} + A\) is invertible under ordinary multiplication.
From the preceding argument, if \(A\) is \(*\)-invertible, then its \(*\)-inverse is
\[ A^{-*} = (I_{n} + A)^{-1} - I_{n}. \]
Indeed, if
\[ B = (I_{n} + A)^{-1} - I_{n}, \]
then
\[ I_{n} + B = (I_{n} + A)^{-1}, \]
and hence
\[ A * B = B * A = 0. \]
One may also rewrite this as
\[ A^{-*} = -(I_{n} + A)^{-1}A. \]
Indeed,
\[\begin{aligned} (I_{n} + A)^{-1} - I_{n} &= (I_{n} + A)^{-1} - (I_{n} + A)^{-1}(I_{n} + A) \\ &= -(I_{n} + A)^{-1}A. \end{aligned}\]
For \(n \ge 2\), the operation \(*\) need not be commutative. It is enough to give an example in \(M_{2}(\mathbb{R})\). Let
\[ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}. \]
Then
\[ AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad BA = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. \]
Thus
\[ AB \ne BA. \]
Now
\[ A * B = A + B + AB = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \]
whereas
\[ B * A = B + A + BA = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}. \]
Therefore
\[ A * B \ne B * A. \]
For \(n > 2\), the same matrices may be embedded in the upper-left \(2 \times 2\) block of \(M_{n}(\mathbb{R})\), with all remaining entries equal to zero. Hence \(*\) is not commutative in general for any \(n \ge 2\).