How to show the existence of a subgroup of an abelian group?? If an abelian group has subgroups of orders m and n, respectively, then show that it has a subgroup whose order is the least common multiple of m and n.
I tried solving this on assumption that if an abelian group has a subgroup of order, say, N then it has an element whose order is also N. This leads me to the point where I arrive at two different elements having orders m and n as in the question, whereby I obtain a third element whose order is the l.c.m of m and n. 
Now, was my assumption correct in the first place?? If it's wrong, please let me know and help me get this solution right.
 A: Let $H_0$ and $H_1$ be subgroups of size $m,n$ and let $H_2=H_0\cup H_1=\{h_1,h_2\dots h_s\}$ and suppose the elements have orders $o_1,o_2\dots o_s$ respectively .Then the set $H_3=\{h_1^{e_1}h_2^{e_2}\dots h_2^{e_s}| e_1,e_2\dots e_s\in \mathbb Z\}$ is a subgroup, because $G$ is abelian. It is also finite as it has at most $o_1o_2\dots o_s$ elements.
By Lagrange's theorem $m$ and $n$ divide $|H_3|$, so $lcm(m,n)$ divides $|H_3|$, so by the following lemma $H$ has a subgroup of order $lcm(m,n)$ (Notice $H_3$ is clearly abelian also).
Lemma: Let $G$ be a finite abelian group and $d$ a divisor of $|G|$, then $G$ has a subgroup of order $d$ (in other words, every abelian group is a converse lagrange group).
Proof:
The proof is by strong induction over $n$
If $|G|=1$ it is trivial. The inductive case is as follows:
If $d=1$ it is trivial, otherwise take $p$ a prime that divides $d$. By cauchy's theorem there is a subgroup $N$ of order $p$, because $G$ is abelian $N$ is normal, so we can consider the canonical projection $\varphi:G\rightarrow \frac{G}{N}$.
Now notice that $d/p$ divides the order of $\frac{G}{N}$, so by the induction hypothesis there is a subgroup $K\leq \frac{G}{N}$ of order $d/p$.
Consider the subgroup $H=\varphi^{-1}(K)$ what is it's order ? Consider the restriction of $\varphi$ to $H$, the kernel is $N$ and the image is $K$, we conclude $\frac{H}{N}\cong K$, so $H$ has order $p\frac{d}{p}=d$.
We have found a subgroup of order $d$.
A: I believe your assumption that our group has an element of order $N$ is not always true- the converse of Lagrange's theorem does not always hold. Also just because an element $g^N=e$ doesnt mean it has order N either (e.g. could have order $N/2$).
Firstly, for my solution it useful to show that 
if $g \in G$ has order $n$, then $g^{k}$ has order $\dfrac{n}{(n,k)}$ for any $k \in \mathbb{Z}$ (1)
First we prove the special case when $m$ and $n$ are positive integers such that $(m,n)=1$:
The required element of $G$ is $xy$ and the least common multiple in this special case where $(m,n)=1$ is $mn$. Let $\left\vert xy \right\vert = c$ then we will show that $c = mn$ as required.
Since $(xy)^{m} = y^{m}$ has order $n$ according to (1) and also we have $n \mid c$. Likewise, applying the same argument, since $(xy)^{n}= x^{n}$ has order $m$ we have $m \mid c$. Therefore, we have $mn \mid c$ because $(m,n)=1$. Furthermore, since $(xy)^{mn} = e$ we have $c \mid mn$ and hence equality $c=mn$, proving the special case where $m$ and $n$ are relatively prime.
Now we are ready to prove the exercise for the general case. Let's reiterate the conditions and statement: Suppose $x,y \in G$, $\left\vert x \right\vert = m$,$\left\vert y \right\vert = n$ and $xy=yx$ then there exists an element in $G$ of order $[m,n]$, the least common multiple of $m$ and $n$. Let's write out $m$ and $n$:
$$m = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$$
$$n = p_{1}^{\beta_{1}}p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}}$$
so $[m,n] = p_{1}^{max\{\alpha_{1},\beta_{1}\}}p_{2}^{max\{\alpha_{2},\beta_{2}\}} \cdots p_{k}^{max\{\alpha_{k},\beta_{k}\}}$. Clearly, $max\{\alpha_{i},\beta_{i}\} = \alpha_{i}$ or $\beta_{i}$.
If $max\{\alpha_{i},\beta_{i}\} = \alpha_{i}$ then by (1) we know the order of $x^{\frac{m}{p_{i}^{\alpha_{i}}}}$ is  $\dfrac{m}{(m,\frac{m}{p_{i}^{\alpha_{i}}})} = p_{i}^{\alpha_{i}}$. Likewise, if $max\{\alpha_{i},\beta_{i}\} = \beta_{i}$ then by (1) we know the order of $y^{\frac{n}{p_{i}^{\beta_{i}}}}$ is  $\dfrac{n}{(n,\frac{n}{p_{i}^{\beta_{i}}})} = p_{i}^{\beta_{i}}$. Therefore, for every $1 \le i \le k$ there exists element $z_{i}$ such that $\left\vert z_{i} \right\vert = p_{i}^{max\{\alpha_{i},\beta_{i}\}}$ where $z_{i}$ is some power of $x$ or $y$ and since $xy=yx$ we have $z_{i}z_{j}=z_{j}z_{i}$ for all $1 \le i \le j \le k$. Hence, by the Special Case, since distinct $z_{i}$ have orders which are mutually relatively prime we have $\left\vert z_{1}z_{2} \cdots z_{k} \right\vert = p_{1}^{max\{\alpha_{1},\beta_{1}\}}p_{2}^{max\{\alpha_{2},\beta_{2}\}} \cdots p_{k}^{max\{\alpha_{k},\beta_{k}\}} = [m,n]$ as desired.
A: Let $U$ and $V$ be the subgroups of order $m$ and $n$ respectively, where the prime decomposition of $m$ and $n$ is given by, 
$$
m = \prod_{i=1}^K p_i^{\alpha_i}\:\: 
\text{and}\:\: 
n = \prod_{j=1}^L p_j^{\beta_j}
$$ 
For any prime $p$  such that $p|m$ or $p|n$ or both, consider the maximum power of $p$ that divides either $m$ or $n$ or both,   let that be $\gamma(p)$. Let $S$ be set of all such primes. Then,
$$lcm(m,n) = \prod _{p \in S} p^{\gamma(p)}$$ 
For any  $ p \in S$, there will be a corrsponding  p-syllow  subgroup either in $U$ or $V$ or both, let that subgroup be $H_p$ and $o(H_p) = p^{\gamma(p)}$. Now the desired subgroup of order $lcm(m,n)$ is direct product of all such $H_p$,
$$H = \prod _{p\in S} H_p$$
