1. (a) Let \(\mathrm{G}\) be an abelian group and \(a, b \in \mathrm{G}\). Show that \((a b)^n=a^n b^n\) for all \(n \in \mathbf{N}\), using the principle of induction.
Answer: To prove the statement using the principle of mathematical induction, we need to first establish the base case and then assume the statement is true for \(n=k\), for some \(k \geq 1\), and then prove it for \(n=k+1\).
Base case \((n=1):\) When \(n=1\), the statement is trivially true as \(a b=a \cdot b\).
Inductive step: Let’s assume that the statement is true for \(n=k\), i.e., \((a b)^k=a^k b^k\).
We want to prove the statement for \(n=k+1\), i.e., \((a b)^{k+1}=a^{k+1} b^{k+1}\).
Using the inductive hypothesis, we can write:
\[
(a b)^{k+1}=(a b)^k \cdot(a b)=a^k b^k \cdot a b=a^k a \cdot b^k b=a^{k+1} b^{k+1}
\]
In the above, we have used the fact that \(G\) is an abelian group and thus \(a\) and \(b\) commute.
So, by the principle of mathematical induction, \((a b)^n=a^n b^n\) for all \(n \in \mathbf{N}\) in an abelian group \(G\).
(b) Is \(x^n-p\) irreducible in \(\mathrm{Q}[x]\) for any prime \(p\) and natural number \(n\) ? Give reasons for your answer.
Answer: The polynomial \(x^n-p\) is irreducible over \(\mathbb{Q}[x]\) for any prime \(p\) and any natural number \(n \geq 1\), except for the case when \(n=1\).
When \(n=1\), the polynomial \(x-p\) is a linear polynomial, which is irreducible.
When \(n>1\), we can use Eisenstein’s criterion for irreducibility, which states that if there exists a prime number \(p\) such that \(p\) divides all the coefficients of the polynomial \(f(x)\) except the leading coefficient, \(p^2\) does not divide the constant term, and the leading coefficient is not divisible by \(p\), then \(f(x)\) is irreducible over \(\mathbb{Q}\). In the case of \(x^n-p\), we can see that the polynomial \(f(x)=x^n-p\) satisfies Eisenstein’s criterion with \(p=p\), so it is irreducible for \(n>1\).
Therefore, \(x^n-p\) is irreducible in \(\mathbb{Q}[x]\) for any prime \(p\) and any natural number \(n\).
(c) Show that every non-abelian group of order 6 is isomorphic to \(\mathrm{D}_6\).
Answer: Let’s denote by \(G\) a non-abelian group of order 6 , and by \(D_6\) the dihedral group of order 6 , which is the group of symmetries of a regular triangle, including both rotations and reflections.
The order of \(G\) is 6 , so by Lagrange’s theorem, the possible orders of elements in \(G\) are 1,2 , 3 , and 6.
Since \(G\) is non-abelian, it cannot be cyclic (as all cyclic groups are abelian), so there is no element of order 6 . We know there is at least one element of order 1 , which is the identity element.
If all other elements had order 2 , then \(G\) would be abelian (consider the result that in a group where all non-identity elements have order 2 , the group operation is commutative), which contradicts the assumption that \(G\) is non-abelian. Therefore, there must be an element of order 3 in \(G\). Let’s denote this element by \(a\). The subgroup generated by \(a\) is then \(\left\{e, a, a^2\right\}\)
Now, since \(|G|=6\) and we have identified a subgroup of order 3 , there must be 3 other elements in the group. Let’s pick one of these and denote it by \(b . b\) cannot have order 1 or 3 (since all elements of these orders are already in \(\langle a\rangle\) ), so \(b\) must have order 2 or 6 . But we’ve already established that there can’t be any elements of order 6 , so \(b\) must have order 2 .
Then \(b^2=e\) and since \(b\) is not in \(\langle a\rangle\), the remaining elements in \(G\) must be \(b, a b, a^2 b\).
The group operation for \(G\) can then be specified as follows:
1. \(a^3=e\)
2. \(b^2=e\)
3. \(b a b=a^2\)
The third equation follows from the fact that \(b\) cannot commute with \(a\) (since \(G\) is nonabelian) and \(b a\) must be one of the elements in the group, and by process of elimination, it can’t be \(e, a, a^2, b\), or \(a b\), so it must be \(a^2 b\).
But this is exactly the presentation of \(D_6\) :
1. \(r^3=e\) (rotation operation)
2. \(s^2=e\) (reflection operation)
3. \(s r s=r^2\) (a reflection followed by a rotation is equivalent to a rotation in the opposite direction followed by a reflection)
By identifying \(a\) with \(r\) and \(b\) with \(s\), we can then establish an isomorphism between \(G\) and \(D_6\). Therefore, any non-abelian group of order 6 is isomorphic to \(D_6\).