Cyclic groups

Prove or disprove that if every proper subgroup of a group \(G\) is cyclic, then \(G\) is a cyclic group.

\(Solution.\) Let's consider \(U(8) = \{1, 3, 5, 7\}\), which has \(4\) proper subgroups \(\{1\}, \{1, 3\}, \{1, 5\}, \{1, 7\}\) that are cyclic groups, but \(U(8)\) is not a cyclic group since the order of \(3, 5, 7\) are 2 instead of 4. \(\blacksquare\)

Prove or disprove that all of generators of \(\mathbb{Z}_{60}\) are prime.

\(Solution.\) In the case \(a\) is a generator of \(\mathbb{Z}_{60}\), \(a\) must satisfy that \(\text{gcd}(a, 60) = 1\). \(60 = 3 * 4 * 5\), so we need to find a number that \[\text{gcd}(a, i) = 1, i = 3, 4, 5.\] \(49\) is the number. Hence the statement are false. \(\blacksquare\)

Prove or disprove that a group with a finite number of subgroups is finite.

\(Solution.\) Suppose \(G\) has a finite number of subgroups, for each subgroup of \(G\), we claim that its order is finite, otherwise we can construct infinite subgroups of \(G\). Specifically, let's suppose \(H \subset G\) and \(|H| = \infty\), if \(\exists h \in H\), \(|\langle h \rangle| = \infty\), \(\{\langle h^i \rangle \mid i = 1, 2, 3 \cdots\}\) is infinite, thus \(G\) has infinite subgroups; else we claim that \(\{\langle h \rangle | h \in H\}\) has infinite elements, thus \(G\) has infinite subgroups. So \(G\) is group that has a finite number of subgroups which have finite elements, hence \(G\) must be finite. \(\blacksquare\)

Let \(a, b \in G\). Prove the following statements.

1. The order of \(a\) is the same as the order of \(a^{-1}\).
2. For all \(g \in G\), \(|a| = |g^{-1}ag|\).
3. The order of \(ab\) is the same as the order of \(ba\).

\(Proof\).

1. Suppose \(|a| = n\), then \((a^{-1})^na^n = a^{-1} \cdots a^{-1}a \cdots a = 1\) so \(|a^{-1}| \big| n\), if \(|a^{-1}| = k < n\), \((a^{-1})^ka^k = 1\), so \(a^k = 1\), conflict. Hence, \(|a| = |a^{-1}|\).
2. \((g^{-1}ag)^n = g^{-1}a^ng\), so \(|g^{-1}ag| \big | |a|\). It is easy to prove that \(|g^{-1}ag| = |a|\).
3. Similar to the order of \(a^{-1}\). \(\blacksquare\)

Let \(p\) and \(q\) be distinct primes. How many generators does \(\mathbb{Z}_{pq}\) have?

\(Solution.\) \(\phi(pq) = pq - (p - 1) - (q - 1)\). Zero is not a generator, so the answer is \(pq - (p - 1) - (q - 1) -1 = (p - 1)(q - 1)\). \(\blacksquare\)

Let \(p\) be prime and \(r\) be a positive integer. How many generators does \(\mathbb{Z}_{p^r}\) have?

\(Solution.\) \(\forall x \in \mathbb{Z}_{p^r}\), if \(\text{gcd}(x, p^r) \not = 1\), \(x \in \{p, 2p, 3p, \cdots , p^2, p^2 + p, \cdots , p^r - p\}\), there are \(p^{r-1} - 1\) elements that don't generate \(\mathbb{Z}_{p^r}\), zero is not a generator, so there are \(p^r - (p^{r - 1} - 1) - 1 = p^r - p^{r - 1}\) generators of \(\mathbb{Z}_{p^r}\). \(\blacksquare\)

Prove that \(\mathbb{Z}_n\) has an even number of generators for \(n > 2\).

\(Proof.\) \(\forall x \in \mathbb{Z}_n\), if \(x\) is a generator, then \(x^{-1} = -x\) is also a generator of \(\mathbb{Z}_n\) because \(|x| = |x^{-1}|\), so every generator \(x\) is in a pair \(\{x, -x\}\), if \(x = -x\), \(2x = 0\) implies \(n \leqslant 2\), conflict with \(n > 2\). Hence, \(\mathbb{Z}_n\) has an even number of generator for \(n > 2\). \(\blacksquare\)

Suppose that \(G\) is a group and let \(a, b \in G\). Prove that if \(|a| = m\) and \(|b| = n\) with \(\text{gcd}(m, n) = 1\), then \(\langle a \rangle \cap \langle b \rangle = \{e\}\).

\(Proof.\) Suppose \(x \in \langle a \rangle \cap \langle b \rangle\), so \(\exists k\) and \(l\) such that \[x = a^k = b^l\] Since \(a^k = b^l\), \(x^m = a^{km} = e = b^{lm}\), so \(n \big | lm\). Similarly, \(m \big | kn\). As \(\text{gcd}(m, n) = 1\), \(n \big | l\), \(m \big | k\), so \(\exists i\), such that \(k = mi\), thus we have \[x = a^k = a^{mi} = e\]. Hence \(\langle a \rangle \cap \langle b \rangle = \{e\}\). \(\blacksquare\)

Prove that group \(G\) has no nontrivial subgroup if and only if \(G = \{1\}\) or \(G\) is a cyclic group of prime order.

\(Proof.\) If \(G\) has no nontrivial subgroup, then \(G = \{1\}\) or \(G = \langle g \rangle, \forall g \not = 1 \in G\). \(G\) has no nontrivial subgroup, if it's a cyclic group, its order must be prime.

It's obvious that if \(G = \{1\}\) or \(G\) is a cyclic group of prime order, \(G\) has no nontrivial subgroup. \(\blacksquare\)