Group Theory
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12 Dedekind groups
Abelian groups have the property that every subgroup is normal. Let’s study all groups that have this property.
The basic example of a Hamilton group is the group \(Q_8\). This justifies the name Hamilton, because Hamilton popularized the quaternions. We’ll see that this is essentially the only kind of example.
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Proof. Let \(a\) be a nontorsion element and \(c\) another element. First, we’ll see \(a \in Z(G)\). Indeed, if \([b,a] \neq 1\), then since it is in \(\langle b \rangle \cap
\langle a \rangle \), \(b\) is nontorsion, and \([b,a] = a^ba^{-1}=a^{-2}\), similarly \([b,a] = b^2\). Thus \(a^{-2} = [b,a]^a = [ab,a]\), so \(ab\) is also nontorsion, and \([ab,a] = abab =
a^2b^2[b^{-1},a^{-1}] = a^4b^2 = a^2\), a contradiction because \(a^2 = a^{-2}\).
Now for any torsion \(b\), \([b,c] = [ab,c] = 1\) as \(ab\) isn’t torsion, so \(G\) is abelian. □
Now in studying Hamilton groups, we observe that on any finite subgroup, every Sylow-\(p\) is normal, meaning it is a product of its Sylow-\(p\) subgroup, taking the inductive limit of this, we see that a Hamilton group is a
product of Hamilton \(p\)-groups, for different \(p\)s.
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Proof. WLOG, we can assume there are two generators, \(a,b\) that don’t commute. Then let \([a,b] = b^{k-1}=a^{1-l}\). By taking a quotient, we can assume \([a,b]^p = 1\).
Now \(ab = b^ka = b^{k-1}ab^l=\dots =ab^{kl}\), so \(kl \equiv 1 \pmod {|b|}\). By symmetry, it is also true for \(a\), and \(kl-1 = k(l-1)+k-1\), so \(l-1\) and \(k-1\) must have the same \(p\)-adic valuation, and
\(|a| = |b|\). □
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Proof. Again assume \(a,b\) don’t commute and are order \(p^{x+1}\), and \([a,b]^p=1\). Then \([a,b] = a^{kp^x} = b^{lp^x}\), \(k,l \in \ZZ /p\ZZ ^\times \). Then since
\([a,b]^p = 1\), and \(\sum _{\ZZ /p\ZZ } i = 0\), \((ab)^p = a^pb^p\). Thus \((a^{-k}b^{l})^{p^x}=1\), which is a contradiction as \([a^{-k}b^{l},b] = [a,b]^{-k}\), so \(p^{x+1} = |b| = |a^{-k}b^l|\leq
p^x\). □
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Proof. Suppose that \(a,b\) don’t commute and have order \(2^k,k\geq 3\). Then since the exponent of \(\ZZ /2^{k}\ZZ \) is \(2^{k-2}\), we have that \([a,b^{2^{k-2}}] =
[a,b]^{2^{k-2}} = 1\). Now if \([a,b]= a^{2^rx} = b^{2^ry}\) for \(x,y\) odd, then \((a^{-x}b^y)^{2^{k-1}} = 1\), a contradiction as it doesn’t commute with \(a\). □
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Proof. An element of order \(2\) is necessarily in the center. Let \(i,j\) be two non-commuting elements, they must be order \(4\). Clearly they generate a subgroup that is \(Q_8\). If
an element \(x\) commutes with \(i\), then \(xj\) doesn’t so \((xj)^2 = -1\), and if \(x\) doesn’t commute, then \(x^2 = -1\). Either way, we see that the quotient \(G/Q_8\) is an elementary abelian \(2\)-group. When
trying to split this short exact sequence \(1 \to Q_8 \to G \to E \to 0\), the only difficulty is if an element in \(E\) lifts to an element \(e\) of order \(4\). But then \(e^2 = -1\). But \(e\) must commute with either \(i,j\)
or \(ij = k\), so WLOG if it commutes with \(i\), then \(ei\) is another lift with \((ei)^2 = 1\). □