Brian Bi
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Return to table of contents for Brian's unofficial solutions to Artin's Algebra

Section 7.3. \(p\)-Groups

Exercise 7.3.1 The action of \(G\) partitions \(S\) into orbits. Since \(|S|\) is not divisible by \(p\), there must be some orbit \(O\) whose size is not divisible by \(p\). But by the orbit-stabilizer theorem, \(|O| \divides |G|\). Therefore \(|O|\) can only be 1, and the sole element of \(O\) is therefore fixed by \(G\).

Exercise 7.3.2 We are given that \(G/Z\) is cyclic, so there is some coset \(xZ\) that generates \(G/Z\). That is, for every \(g \in G\), there is some \(n\) such that \(g \in (xZ)^n\). Now \((xZ)^n\) is just the coset \(x^n Z\), so \(g = x^n z\) for some \(z \in Z\). For every \(g' \in G\), we can likewise write \(g' = x^{n'} z'\) for some \(n'\) and some \(z' \in Z\). By inspection, \(g\) and \(g'\) commute. We conclude that \(G\) is abelian.

Exercise 7.3.3

  1. \(|Z|\) must divide \(p^3\). It cannot be 1 according to Proposition 7.3.1. It also cannot be \(p^3\) because we are given that \(G\) is nonabelian. The case \(|Z| = p^2\) is ruled out by the Lemma from Section 7.2 exercises since all remaining conjugacy classes (besides those of the central elements) would have to have size \(p\). Therefore the only possible order of the centre is \(p\).
  2. Since the order of the centre is \(p\), the Lemma rules out the existence of any conjugacy classes of size \(p^2\). So all non-central elements belong to a conjugacy class of size \(p\), for which the centralizer has order \(p^2\).
  3. The class equation can only be \(p \times 1 + (p^2 - 1) \times p\) by the above reasoning, where \(m \times n\) denotes \(m\) conjugacy classes of size \(n\).

Exercise 7.3.4 All elements of \(G\) other than the identity must have order 2, 4, or 8. Choose some \(x \in G\) of maximal order and let \(H = \langle x \rangle\). The classification is by cases:

  1. \(x\) has order 8. Thus, \(G \cong C_8\).
  2. \(x\) has order 4. By Exercise 2.8.10, \(H\) is normal in \(G\). Then,
    1. Suppose there exists \(y \in G \setminus H\) such that \([x, y] = 1\). If \(y^2 = x^2\), then set \(y' = yx^{-1}\) so that \(y'^2 = y^2 x^{-2} = x^2 x^{-2} = 1\) and use \(y'\) in place of \(y\). Let \(K = \langle y \rangle\); now we know that \(H \cap K = \{1\}\). By Proposition 2.11.4(a), \(HK\) consists of \(4|K|\) distinct elements. Therefore \(|K| = 2, HK = G\), and the elements \(x\) and \(y\) generate \(G\). By Proposition 2.11.4(b), \(G \cong H \times K \cong C_4 \times C_2\).
    2. Otherwise, suppose there exists \(y \in G \setminus H\) with order 4. Then \(y^3 = y^{-1}\) also isn't in \(H\). The element \(xy\) can't be in \(H\) or \(\langle y \rangle\), and neither can \(yx\), which is also distinct from \(xy\). So \(G = \{1, x, x^2, x^3, y, y^3, xy, yx\}\). Now \(yxy^{-1}\) lies in \(H\) since \(H\) is normal, and it must have the same order as \(x\), so it can only be \(x^3\). This completely determines the multiplication table of \(G\). Now the quaternion group \(Q_8\) has the aforementioned properties with \(x = i, y = j\). Therefore \(G \cong Q_8\).
    3. Otherwise, all remaining elements of \(G \setminus H\) have order 2. Let \(y\) be such an element and let \(K = \{1, y\}\). By Proposition 2.11.4(a), the set \(HK\) consists of 8 distinct elements, so \(HK = G\) as sets. That is, \(G = \{1, x, x^2, x^3, y, xy, x^2 y, x^3 y\}\). Since \(H\) is normal and since \(yxy^{-1}\) has the same order as \(x\), we must have \(yx = xy^3\). This completely determines the multiplication table of \(G\). Now the dihedral group \(D_4\) has the aforementioned properties with \(x = \rho, y = r\). Therefore \(G \cong D_4\).
  3. \(x\) has order 2. Since all non-identity elements have order 2, \(G\) is abelian. Pick \(y \in G \setminus H\). Then we have a subgroup \(I = \{1, x, y, xy\}\) and it is easy to see that \(I \cong C_2 \times C_2\). Pick \(z \in G \setminus I\). Then, by Proposition 2.11.4(c), \(I\langle z\rangle\) is a subgroup of \(G\) which is larger than \(I\), therefore it is equal to \(G\). By Proposition 2.11.4(d), \(G \cong I \times \langle z\rangle \cong C_2 \times C_2 \times C_2\).

There are thus a total of five isomorphism classes for groups of order 8, namely \(C_8, C_2 \times C_4, C_2 \times C_2 \times C_2, D_4, Q_8\). (The division into cases makes it clear that these five groups are pairwise nonisomorphic.)