Applications and further properties

We wish to show, for our first application, that a cyclic group of order $n$ can be written as a direct product of cyclic groups of prime power order.

Theorem 9.2.1   If $G$ is a cyclic group of order $n=\prod_{i=1}^k p_i^{\alpha_i}$, where the $p_i$ are distinct primes and $\alpha_i\geq 1$ are integers, then $G$ is a direct product of cyclic groups of orders $p_i^{\alpha_i}$, $i = 1, 2, ..., k$.

Proof: We designate by $G_i$ the unique cyclic subgroup of order $p_i^{\alpha_i}$ of $G$ (see Theorem 5.2.1) and let $H = G_1G_2...G_k$. $H$ is, of course a subgroup, since the $G_i$ commute, because $G$ is cyclic (and thus, of course, abelian). Also $G_i\subset H$ for every $i = 1, 2, ..., k$; therefore $p_i^{\alpha_i}\vert \, \vert H\vert$, for every $i = 1, 2, ..., k$. Thus $n = lcm (p_i^{\alpha_i})$ (see Theorem 1.2.11, generalized from $2$ factors to $k$ factors) and so $n\vert\, \vert H\vert$. Since $H \leq G$, $\vert H\vert\, \vert G\vert=n$ and so $n= \vert H\vert$. This proves $G = H = G_1G_2...G_k$.

Next, we designate by $H_i$ the cyclic subgroup of $G$ of order $n_i=n/p_i^{\alpha_i}$, $i = 1, 2, ..., k$, and let $W_i = H_i \cap G_i$. Then $W_i \leq H_i$ and also $W_i \leq G_i$. Thus $\vert W_i\vert\, \vert n_i$ and $W_i\vert p_i^{\alpha_i}$, but $gcd(n_i,p_i^{\alpha_i})=1$, so $\vert W_i\vert = 1$. Hence $W_i = H_i \cap G_i = \{e\}$. However, $p_i^{\alpha_i}\vert\, \vert H_i\vert$ for all $j \not= i$. Thus $G_j\subset H_i$ for $j \not= i$ since $H_i$, being cyclic, contains a subgroup of order $p_j^{\alpha_j}$ and that subgroup must be $G_j \leq H_i \leq G$ by uniqueness (see Theorem 5.2.1), so $G_1...G_{i- 1}G_{i+1}...G_k\subset H_i$ and therefore

$$(G_1...G_{i- 1}G_{i+1}...G_k) \cap G_i = \{e\},$$

for all $i = 1, 2, ..., k$. $\Box$

We next prove an important property of direct products.

Theorem 9.2.2   If $G = G_1 \times G_2$, then $G_2 \cong G/G_1$ and $G_1 \cong G/G_2$.

Proof: The theorem is an immediate consequence of the second isomorphism theorem (Theorem 8.3.7). Namely, we have

and since $G_1$ and $G_2$ are both normal in $G$, Theorem 8.3.7 gives $G/G_1 \cong G_2/\{e\} \cong G_2$ and $G/G_2 \cong G_1$. $\Box$

We have already seen one instance (Theorem 8.1.5) in which a normal subgroup $N_1$ of a normal subgroup $N_2$ of a group $G$ is normal in $G$, i.e., $N_1\lhd N_2$, $N_2\lhd G$, and $N_1\lhd G$. The following theorem gives another important case in which this true.

Theorem 9.2.3   If $H$ is a direct factor of the group $G$ (i.e., $G = H \times N$, for some $N \leq G$), then every normal subgroup of $H$ is normal in $G$.

Proof: Let $W \lhd H$. Now by hypothesis, $G = H \times N$. Thus if $g$ is an arbitrary element of $G$, then we can write $g$ in the form $g = hn$, where $h\in H$ and $n \in N$. Now

$$gWg^{-1} = (hn)Wn^{-1}h^{-1} = hWh^{-1} = W,$$

where we have used the fact that elements of $H$ and $N$ commute (see Theorem 9.1.4). Since $g$ was an arbitrary element of $G$, we see indeed that $W \lhd G$. $\Box$

We continue with another application of direct products to cyclic groups.

Theorem 9.2.4   Let $C_m$ and $C_n$ be cyclic groups of orders $m$ and $n$, respectively. $G = C_m \times C_n$ is cyclic if and only if $gcd (m,n) = 1$.

Proof: Suppose $gcd (m,n ) > 1$. We shall show that $G$ is not cyclic. Let $p$ be a prime such that $p\vert\,\vert\, gcd (m,n)$, so $p\vert m$ and $p\vert n$. Thus $C_m$ has a cyclic subgroup of order $p$ and $C_n$ has also a cyclic subgroup of order $p$ (see Theorem 5.2.1), but $C_m \cap C_n = \{e\}$. Consequently, $G$ has at least two cyclic subgroups of order $p$. This implies, by Theorem 5.2.1, that $G$ is not cyclic.

Next suppose that $gcd (m,n) = 1$ and let $C_m = \langle a\rangle$ and $C_n = \langle b\rangle$. Then $ab \in G = C_m \times C_n$ and $o(ab) = mn$, for if we let $t = o(ab)$, then

$$(ab)^{mn} = a^{mn}b^{mn} = e$$

so $t\vert mn$. Since

$$e = (ab)^{mt} = a^{mt}b^{mt} = b^{mt},$$

so $n\vert mt$. But $gcd (m,n) = 1$, so $n\vert t$. Similarly, we can show that $m\vert t$ and, again, since $gcd (m,n) = 1$, we must have $lcm (m,n)= mn\vert t$. Thus $mn = t = o(ab)$. However, $\vert G\vert = mn$ so $G = \langle ab\rangle$. $\Box$

As our last application, we prove that the Euler $\phi$-function is multiplicative, i.e., if $gcd (m,n) = 1$, then $\phi(mn) = \phi(m)\phi(n)$. Suppose $C_m$ is a cyclic group of order $m$, and $C_n$ is a cyclic group of order $n$, where $gcd (m,n) = 1$. Then $C_m \times C_n = C_{mn}$, is a cyclic group of order $mn$ by Theorem 9.2.4. We know from our discussion of cyclic groups in Chapter 5 (in particular see Corollary 5.2.3) that $C_m$ has $\phi(m)$ generators and that $C_n$ has $\phi(n)$ generators. While $C_{mn}$ has $\phi (mn)$ generators. However, it is easy to see (see exercise 2 for this section) that every generator of $C_{mn}$ must be of the form $(a,b)$ where $a$ is a generator of $C_m$ and $b$ is a generator of $C_n$. Thus we have proven the following result.

Theorem 9.2.5   If $m,n \in \mathbb{N}$ and $gcd (m,n) = 1$, then $\phi(mn) = \phi(m)\phi(n)$. In other words, $\phi$ is multiplicative.