Problem on Finite Abelian Groups: Part-I
Welcome to our comprehensive guide on finite abelian groups! In this article, you will discover an array of thoughtfully chosen problems along with detailed, step-by-step solutions. Our exploration emphasizes key concepts such as group isomorphism, primary decomposition, and the Chinese Remainder Theorem. Whether you are a student seeking clarity or a mathematics enthusiast eager to deepen your understanding of group theory, these examples are designed to engage and inspire. Embrace the journey to master finite abelian groups and enhance your mathematical skills!
Problem-1
Problem:
Prove that
\(\mathbb{Z}_{8} \oplus \mathbb{Z}_{2}\)
is not isomorphic to
\(\mathbb{Z}_{4} \oplus \mathbb{Z}_{4}\).
Answer:
Observe that \(\mathbb{Z}_{8} \oplus \mathbb{Z}_{2}\) can be written as \(\mathbb{Z}_{2^3} \oplus \mathbb{Z}_{2^1}\) with invariant factors \((3, 1)\). In contrast, \(\mathbb{Z}_{4} \oplus \mathbb{Z}_{4}\) is equivalent to \(\mathbb{Z}_{2^2} \oplus \mathbb{Z}_{2^2}\) and has invariant factors \((2, 2)\). Since the tuples \((3, 1)\) and \((2, 2)\) differ, the groups are not isomorphic.
Problem-2
Problem:
Determine whether the groups
\(\mathbb{Z}_{5} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{3}\)
and
\(\mathbb{Z}_{5} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{9}\)
are isomorphic.
Answer:
Begin by decomposing each group into its prime components. The first group contains two copies of \(\mathbb{Z}_{5}\) (its 5‑part) and two copies of \(\mathbb{Z}_{3}\) (its 3‑part). The second group also has the 5‑part as \(\mathbb{Z}_{5} \oplus \mathbb{Z}_{5}\), but its 3‑part is given by the cyclic group \(\mathbb{Z}_{9}\) (since \(9=3^2\)). Notice that while \(\mathbb{Z}_{3} \oplus \mathbb{Z}_{3}\) is non-cyclic, \(\mathbb{Z}_{9}\) is cyclic. This difference in cyclicity shows that the groups are not isomorphic.
Problem-3
Problem:
Assess whether
\(\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2}\)
is isomorphic to
\(\mathbb{Z}_{4} \oplus \mathbb{Z}_{2}\).
Answer:
The group \(\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2}\) is an elementary abelian 2‑group where every nonzero element has order 2. On the other hand, \(\mathbb{Z}_{4} \oplus \mathbb{Z}_{2}\) contains at least one element of order 4 (for example, the element \((1,0)\)). Since one group contains an element of higher order than the other, they are not isomorphic.
Problem-4
Problem:
Show that
\(\mathbb{Z}_{12}\)
is isomorphic to
\(\mathbb{Z}_{3} \oplus \mathbb{Z}_{4}\).
Answer:
Recognize that \(12 = 3 \times 4\) and the numbers 3 and 4 are relatively prime. By the Chinese Remainder Theorem, we deduce the natural isomorphism:
\[
\mathbb{Z}_{12} \cong \mathbb{Z}_3 \oplus \mathbb{Z}_4.
\]
This confirms the isomorphism between the two groups.
Problem-5
Problem:
Determine if
\(\mathbb{Z}_{6} \oplus \mathbb{Z}_{6}\)
is isomorphic to
\(\mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{6}\).
Answer:
Note that \(\mathbb{Z}_6\) is isomorphic to \(\mathbb{Z}_2 \oplus \mathbb{Z}_3\). Therefore, we can express:
\[
\mathbb{Z}_6 \oplus \mathbb{Z}_6 \cong (\mathbb{Z}_2 \oplus \mathbb{Z}_3) \oplus (\mathbb{Z}_2 \oplus \mathbb{Z}_3) \cong (\mathbb{Z}_2 \oplus \mathbb{Z}_2) \oplus (\mathbb{Z}_3 \oplus \mathbb{Z}_3).
\]
Similarly, rewriting the second group gives:
\[
\mathbb{Z}_3 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_6 \cong \mathbb{Z}_3 \oplus \mathbb{Z}_2 \oplus (\mathbb{Z}_2 \oplus \mathbb{Z}_3) \cong (\mathbb{Z}_2 \oplus \mathbb{Z}_2) \oplus (\mathbb{Z}_3 \oplus \mathbb{Z}_3).
\]
Since both groups decompose into identical components, they are isomorphic.
Problem-6
Problem:
Verify whether
\(\mathbb{Z}_{15}\)
is isomorphic to
\(\mathbb{Z}_{3} \oplus \mathbb{Z}_{5}\).
Answer:
Since \(15 = 3 \times 5\) and 3 and 5 are coprime, the Chinese Remainder Theorem guarantees that:
\[
\mathbb{Z}_{15} \cong \mathbb{Z}_3 \oplus \mathbb{Z}_5.
\]
Thus, the two groups are isomorphic.
Problem-7
Problem:
Demonstrate that
\(\mathbb{Z}_{2} \oplus \mathbb{Z}_{4}\)
is not isomorphic to
\(\mathbb{Z}_{8}\).
Answer:
Although both groups have order 8, a closer look at the orders of their elements reveals a critical difference. The cyclic group \(\mathbb{Z}_8\) has an element of order 8, whereas the maximum element order in \(\mathbb{Z}_{2} \oplus \mathbb{Z}_{4}\) is determined by \(\operatorname{lcm}(2, 4) = 4\). Hence, lacking an element of order 8, the groups cannot be isomorphic.
Problem-8
Problem:
Determine if
\(\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3}\)
is isomorphic to
\(\mathbb{Z}_{6} \oplus \mathbb{Z}_{2}\).
Answer:
Both groups have order 12. Decomposing into primary components, observe that for \(\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3}\), the 2‑part is \(\mathbb{Z}_{2} \oplus \mathbb{Z}_{2}\) and the 3‑part is \(\mathbb{Z}_{3}\). Recognizing that \(\mathbb{Z}_6 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3\), we can express:
\[
\mathbb{Z}_{6} \oplus \mathbb{Z}_{2} \cong (\mathbb{Z}_2 \oplus \mathbb{Z}_3) \oplus \mathbb{Z}_2 \cong (\mathbb{Z}_2 \oplus \mathbb{Z}_2) \oplus \mathbb{Z}_{3}.
\]
Since both decompositions are identical, the groups are isomorphic.
Problem-9
Problem:
Ascertain whether
\(\mathbb{Z}_{9} \oplus \mathbb{Z}_{3}\)
is isomorphic to
\(\mathbb{Z}_{27}\).
Answer:
Although both groups have an order of 27, their structures diverge significantly. In \(\mathbb{Z}_9 \oplus \mathbb{Z}_3\), the maximum order of an element is \(\operatorname{lcm}(9, 3) = 9\), which means the group is not cyclic. In contrast, \(\mathbb{Z}_{27}\) is cyclic and contains an element of order 27. Because a cyclic group cannot be isomorphic to a non-cyclic group, these groups are not isomorphic.
Problem-10
Problem:
Prove that
\(\mathbb{Z}_{4} \oplus \mathbb{Z}_{6}\)
is isomorphic to
\(\mathbb{Z}_{2} \oplus \mathbb{Z}_{12}\).
Answer:
Start by expressing \(\mathbb{Z}_6\) as \(\mathbb{Z}_2 \oplus \mathbb{Z}_3\) (since \(6 = 2 \times 3\) with \(\gcd(2,3)=1\)). Then:
\[
\mathbb{Z}_4 \oplus \mathbb{Z}_6 \cong \mathbb{Z}_4 \oplus (\mathbb{Z}_2 \oplus \mathbb{Z}_3) \cong (\mathbb{Z}_4 \oplus \mathbb{Z}_2) \oplus \mathbb{Z}_3.
\]
Notice that by the Chinese Remainder Theorem, \(\mathbb{Z}_{12} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_3\). Hence,
\[
\mathbb{Z}_2 \oplus \mathbb{Z}_{12} \cong \mathbb{Z}_2 \oplus (\mathbb{Z}_4 \oplus \mathbb{Z}_3) \cong (\mathbb{Z}_2 \oplus \mathbb{Z}_4) \oplus \mathbb{Z}_3.
\]
Since both groups share the same decomposition, they are isomorphic.
Conclusion
In this problem set on finite abelian groups, we have methodically dissected each scenario by examining invariant factors, primary components, and element orders. These examples illustrate the critical role of the fundamental theorem of finitely generated abelian groups in understanding group isomorphisms. Continue practicing these techniques, and let your journey in group theory inspire further exploration and mastery. Happy problem solving!
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FAQs
Inner Product Space
- What is an inner product space?
An inner product space is a vector space equipped with an additional structure called an inner product. The inner product allows for the definition of geometric concepts such as length, angle, and orthogonality.
- What is an inner product?
An inner product is a function that takes two vectors from the vector space and returns a scalar, typically denoted as ( langle u, v rangle ) for vectors ( u ) and ( v ). This function must satisfy certain properties: linearity in the first argument, symmetry, and positive-definiteness.
- What are the properties of an inner product?
- Linearity in the first argument:** ( langle au + bv, w rangle = a langle u, w rangle + b langle v, w rangle ) for all scalars ( a, b ) and vectors ( u, v, w ).
- Symmetry:** ( langle u, v rangle = langle v, u rangle ) for all vectors ( u, v ).
- Positive-definiteness:** ( langle u, u rangle geq 0 ) for all vectors ( u ), and ( langle u, u rangle = 0 ) if and only if ( u ) is the zero vector.
- How does the inner product relate to the norm of a vector?
The norm (or length) of a vector ( u ) in an inner product space is defined as the square root of the inner product of the vector with itself, i.e., ( |u| = sqrt{langle u, u rangle} ).
- What is orthogonality in an inner product space?
Two vectors ( u ) and ( v ) are orthogonal if their inner product is zero, i.e., ( langle u, v rangle = 0 ). Orthogonality generalizes the concept of perpendicularity in Euclidean space.
- What is the Cauchy-Schwarz inequality?
The Cauchy-Schwarz inequality states that for all vectors ( u ) and ( v ) in an inner product space, ( |langle u, v rangle| leq |u| |v| ). This inequality is fundamental in the study of inner product spaces.
- What is an orthonormal basis?
An orthonormal basis of an inner product space is a basis consisting of vectors that are all orthogonal to each other and each have unit norm. This means that for an orthonormal basis ( {e_1, e_2, ldots, e_n} ), ( langle e_i, e_j rangle = 1 ) if ( i = j ) and ( 0 ) otherwise.
- How do you project a vector onto another vector in an inner product space?
The projection of a vector ( u ) onto a vector ( v ) is given by ( left(frac{langle u, v rangle}{langle v, v rangle}right) v ). This formula uses the inner product to find the scalar component of ( u ) in the direction of ( v ).
- What is the Gram-Schmidt process?
The Gram-Schmidt process is a method for orthonormalizing a set of vectors in an inner product space. Given a set of linearly independent vectors, the process constructs an orthonormal set of vectors that spans the same subspace.
