Fraleigh Abstract algebra exercise 6.56 (b)

Question

Let $G$ be an abelian group and let $H$ and $K$ be finite cyclic subgroups with $|H|=r$ and $|K|=s$.($|\cdot|$ imply order of group $\cdot$,$<a>$ imply cyclic group generated by $a$ ).

(a) Show that if $r$ and $s$ are relatively prime, then $G$ contains a cyclic subgroup of order $rs$.

(b) Generalizing part (a), show that $G$ contains a cyclic subgroup of order the least common multiple of $r$ and $s$.

I proved part (a), and solution manual suggest the proof of (b) using part (a) like following.

(b) Let $d$ be the gcd of $r$ and $s$, and let $s = dq$ so that $q$ and $r$ are relatively prime and $rq = rs/d$ is the least common multiple of $r$ and $s$. Let $a$ and $b$ be generators of $H$ and $K$ respectively. Then $|<a>| = r$ and $|<b^d>| = q$ where $r$ and $q$ are relatively prime. Part(a) shows that the element $ab^d$ generates a cyclic subgroup of order $rq$ which is the least common multiple of $r$ and $s$.

However when $r=2^2*3,s=2*3^2$,$d=2*3$ and $q=3$ and $q$ and $r$ is not relatively prime. In this case, solution can't explain the problem. Is there anyone who know the exact proof for (b).

Answer 1

I agree the hint in the book doesn't seem to be complete.

Write $r = p_1^{r_1} \cdots p_m^{r_m}$ and $s = p_1^{s_1} \cdots p_m^{s_m}$, where the $p_i$ are prime and $r_i, s_i$ are integers $\geq 0$. Reorder the primes $p$ such that there exists $1 \leq t \leq m$ such that $r_i \geq s_i$ iff $i \leq t$. That is, $\gcd(r, s) = p_1^{r_1} \cdots p_t^{r_t} p_{t+1}^{s_{t+1}} \cdots p_m^{s_m}$.

Then consider $f = a^{p_{t+1}^{r_{t+1}} \cdots p_m^{r_m}}$ and $g = b^{p_1^{s_1} \cdots p_t^{s_t}}$. Then $f$ has order $p_1^{r_1} \cdots p_t^{r_t}$, $g$ has order $p_{t+1}^{s_{t+1}} \cdots p_m^{s_m}$, and you can apply part (a) to $f$ and $g$.