June, 2013
ELECTIVE COURSE : MATHEMATICS
MTE-06 : ABSTRACT ALGEBRA
1. (a) Prove that any finite group is a subgroup of a permutation group.
Answer:
To prove this we will use Cayley’s theorem.
Cayley’s theorem states that every finite group \(G\) is isomorphic to a subgroup of the symmetric group \(S_n\), where \(n\) is the order of the group \(G\) (i.e., the number of elements in \(G\) ). The symmetric group \(S_n\) is the group of all permutations of \(n\) elements.
Proof:
1. Let \(G\) be a finite group of order \(n\). We will construct an injective (one-to-one) group homomorphism from \(G\) to the symmetric group \(S_n\).
2. For each element \(g \in G\), define the function \(L_g: G \rightarrow G\) as \(L_g(x)=g x\) for all \(x \in G\) . The function \(L_g\) is a bijection, as it has an inverse given by \(L_{g^{-1}}\).
3. Define a function \(\phi: G \rightarrow S_n\) by \(\phi(g)=L_g\). We will show that \(\phi\) is an injective group homomorphism.
4. First, let’s show that \(\phi\) is a homomorphism. Let \(g_1, g_2 \in G\), then
\[
\phi\left(g_1 g_2\right)(x)=L_{g_1 g_2}(x)=\left(g_1 g_2\right) x=g_1\left(g_2 x\right)=L_{g_1}\left(L_{g_2}(x)\right)=\left(L_{g_1} \circ L_{g_2}\right)(x)
\]
Since this holds for all \(x \in G\), we have \(\phi\left(g_1 g_2\right)=\phi\left(g_1\right) \circ \phi\left(g_2\right)\), which shows that \(\phi\) is a homomorphism.
5. Now, let’s show that \(\phi\) is injective. Suppose \(\phi\left(g_1\right)=\phi\left(g_2\right)\) for some \(g_1, g_2 \in G\). Then,
\[
L_{g_1}=\phi\left(g_1\right)=\phi\left(g_2\right)=L_{g_2}
\]
Since \(L_{g_1}(e)=g_1\) and \(L_{g_2}(e)=g_2\), where \(e\) is the identity element of \(G\), it follows that \(g_1=g_2\), which shows that \(\phi\) is injective.
6. Therefore, \(\phi(G)\) is an isomorphic copy of \(G\) inside \(S_n\), which implies that \(G\) is a subgroup of a permutation group.
Hence, by Cayley’s theorem, we have proved that any finite group is a subgroup of a permutation group.
(b) Let \(d\) be a Euclidean norm on a Euclidean domain D. Show that if \(s \in Z\) such that \(s+d(1)>0\), then \(g \quad: \quad D \backslash\{0\} \rightarrow Z \quad\) : \(g(a)=d(a)+s\), for non-zero \(a \in D\), is a Euclidean norm on D.
Answer:
To show that \(g: D \backslash\{0\} \rightarrow \mathbb{Z}\) defined by \(g(a)=d(a)+s\) for non-zero \(a \in D\) is a Euclidean norm on \(D\), we need to verify the two properties of Euclidean norms:
1. \(g(a) \geq 0\) and \(g(a)=0\) if and only if \(a=0\).
2. For any two elements \(a, b \in D\) with \(b \neq 0\), there exist \(q, r \in D\) such that \(a=b q+r\) and either \(r=0\) or \(g(r)<g(b)\).
First, we know that \(d\) is a Euclidean norm on \(D\). Therefore, it satisfies these properties. Let’s now check if the function \(g\) satisfies these properties:
1. Since \(d(a) \geq 0\) for all non-zero \(a \in D\), and given that \(s+d(1)>0\), we have \(g(a)=\) \(d(a)+s \geq 0\) for all non-zero \(a \in D\). Moreover, \(g(a)=0\) implies that \(d(a)+s=0\), which is a contradiction because \(d(a) \geq 0\) and \(s+d(1)>0\). Thus, \(g(a)=0\) if and only if \(a=0\).
2. Let \(a, b \in D\) with \(b \neq 0\). Since \(d\) is a Euclidean norm on \(D\), there exist \(q, r \in D\) such that \(a=b q+r\) and either \(r=0\) or \(d(r)<d(b)\). If \(r=0\), then we are done. If \(r \neq 0\), then
\[
g(r)=d(r)+s<d(b)+s=g(b)
\]
Thus, in either case, there exist \(q, r \in D\) such that \(a=b q+r\) and either \(r=0\) or \(g(r)<\) \(g(b)\)
Since the function \(g\) satisfies both properties of Euclidean norms, it is a Euclidean norm on D.