Chapter 9: Problem 42
Prove that \(\operatorname{Inn}(G)\) is a subgroup of \(\operatorname{Aut}(G)\).
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Chapter 9: Problem 42
Prove that \(\operatorname{Inn}(G)\) is a subgroup of \(\operatorname{Aut}(G)\).
These are the key concepts you need to understand to accurately answer the question.
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Find two nonisomorphic groups \(G\) and \(H\) such that \(\operatorname{Aut}(G) \cong \operatorname{Aut}(H)\).
Find the order of each of the following elements. (a) (3,4) in \(\mathbb{Z}_{4} \times \mathbb{Z}_{6}\) (b) (6,15,4) in \(\mathbb{Z}_{30} \times \mathbb{Z}_{45} \times \mathbb{Z}_{24}\) (c) (5,10,15) in \(\mathbb{Z}_{25} \times \mathbb{Z}_{25} \times \mathbb{Z}_{25}\) (d) (8,8,8) in \(\mathbb{Z}_{10} \times \mathbb{Z}_{24} \times \mathbb{Z}_{80}\)
Prove that \(U(8)\) is isomorphic to the group of matrices $$\left(\begin{array}{ll}1 & 0 \\\0 & 1\end{array}\right),\left(\begin{array}{cc}1 & 0 \\\0 & -1 \end{array}\right),\left(\begin{array}{cc}-1 & 0 \\\0 & 1 \end{array}\right),\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)$$
Prove that \(S_{3} \times \mathbb{Z}_{2}\) is isomorphic to \(D_{6}\). Can you make a conjecture about \(D_{2 n}\) ? Prove your conjecture.
Groups of order \(2 p .\) In this series of exercises we will classify all groups of order \(2 p,\) where \(p\) is an odd prime. (a) Assume \(G\) is a group of order \(2 p,\) where \(p\) is an odd prime. If \(a \in G,\) show that \(a\) must have order \(1,2, p,\) or \(2 p\). (b) Suppose that \(G\) has an element of order \(2 p\). Prove that \(G\) is isomorphic to \(\mathbb{Z}_{2 p}\). Hence, \(G\) is cyclic. (c) Suppose that \(G\) does not contain an element of order \(2 p .\) Show that \(G\) must contain an element of order \(p .\) Hint: Assume that \(G\) does not contain an element of order \(p\). (d) Suppose that \(G\) does not contain an element of order \(2 p .\) Show that \(G\) must contain an element of order 2 .(e) Let \(P\) be a subgroup of \(G\) with order \(p\) and \(y \in G\) have order 2 . Show that \(y P=P y\). (f) Suppose that \(G\) does not contain an element of order \(2 p\) and \(P=\langle z\rangle\) is a subgroup of order \(p\) generated by \(z\). If \(y\) is an element of order 2 , then \(y z=z^{k} y\) for some \(2 \leq k
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