Subgroup of order 3 
Prove or disprove that every abelian group of order divisible by $3$ contains a subgroup of order $3$. 

I think you're meant to solve it without using Sylow Theorems or Abelian Group Classifications. It appeared in the book before these theorems are introduced!
 A: I would do the following:


*

*Show that every abelian group of order divisible by $3$ has an element $a$ whose order is divisible by $3$. You can do this by induction: Take $a \in G$. If the order of $a$ is not divisible by $3$, then the quotient group $G/\langle a \rangle$ must have order divisible by $3$. $\langle a \rangle$ is the group generated by $a$. Now by induction, $G/\langle a \rangle$ has an element $b$ whose order is divisible by $3$. Show that, if you pick a representative $b' \in G$ of $b$, the order of $b'$ must be divisible by $3$.

*Now, if $a$ has order equal to $3k$, show that $a^k$ has order $3$ and therefore generates a cyclic subgroup of order $3$.
A: Let $G$ have order divisible by $3$. Let $H$ be a subgroup of $G$ maximal among those with order not divisible by $3$, and let $x \in G - H$. 
Write $K$ for the subgroup generated by $H$ and $x$. Then $K/H$ is generated by $\bar{x}$, so $K/H$ is cyclic. Moreover, $K/H$ must have order divisible by $3$, and this is the order of $\bar{x}$. This proves in turn that $x$ has order a multiple of $3$. The cyclic group generated by $x$ therefore has a subgroup of order $3$.
A: I am working on same book and the book has not covered quotient groups by this point. A proof I came up with that only uses the concepts explained so far in the book is shown below. I hope it makes sense.
Form the cyclic subgroup generated by each element of the group. You now have a list of cyclic groups. If one group on the list is a subset of another remove it from the list. The remaining cyclic groups do not intersect except for the identity.
If 3 divides the order of $<g>$. Then $|g|$=3$x$ for some $x$. Therefore |$g^x$| is 3 and $g^x$ would generate a group of order 3. So if the theorem is false none of the cyclic groups in the list formed above have order divisible by 3. 
However if we take the internal direct product of the groups in the list it is isomorphic to the whole group. You should be able to see it satisfies the conditions for the internal direct product since it would contain every product of every element in the group, the intersection of the groups is the identity and the elements all commute.
However for an internal direct product the order of the group is the product of the orders of the groups it is made of, but we said earlier that if the theorem is false then 3 doesn't divide the order of any of these groups. But we know 3 does divide the order of the main group. 
If some numbers multiply together to make a number that is divisible by a prime then at least one of the numbers must be divisible by that prime. Therefore at least one of the groups must have an order divisible by 3 for the orders to multiply together to make the order of the main group, which is divisible by 3. Therefore there is an element of order 3 and it generates a group of order 3.
This proof works for any prime number, not just 3, so this is a proof of Cauchy's theorem but only for abelian groups.
A: I am at the same chapter. I do not know about Quotient Groups. This is how I solved it, or I think I solved it.
Let G be a group of order 3n. Since the statement is trivially true for n=1, let us use induction and assume that the statement is true for some $n\in Z_{+}$. Let $G_{1}$ be the group whose order is greater than |G| and is a multiple of 3, and there is no group whose order is a multiple of 3 and greater than |G| and less than |$G_{1}$|. Let $3n_{1}=|G_{1}|$.
If $G_{1}$ has any subgroup of order that is a multiple of 3, then by induction, we are done. Let us then assume that it does not. This means $G_{1}$ is not cyclic. Then let $H$ be a proper subgroup of $G_{1}$. Such a subgroup can be found since $G_{1}$ is not cyclic. Then let $a\in G_{1}-H.$ Let $ d$ be the smallest positive integer such that $a^{d}\in H$. Then $|a|=md$ for some positive integer $m$, by division algorithm.
Consider the cosets $H, aH, a^{2}H, ..., a^{d-1}H$. $a^{i}H\neq a^{j}H$ if $i\neq j,$ $i,j\in$ {$1,..,d-1$}. These are distinct cosets and their union $H_{1}$ forms a subgroup of order $d|H|$. If $d$ is not divisible by 3, we repeat this process with $H_{1}$ until we get a subgroup $H_{k}$ such that $|H_{k}|=dd_{1}d_{2}..d_{k-1}|H|$. Then some $d_{i}$ is divisible by 3. So, we have an element $b$ whose order is divisible by 3, thus we can get a subgroup of order 3 which is the cyclic group $<b^{\frac{q}{3}}>$, where $|b|=q$.
Thus our induction hypothesis is satisfied.
We can also avoid induction and use the method of constructing subgroups of the given group $G$ whose order is divisible by 3, starting from a subgroup $H$ where $H$ is a cyclic group whose order is not divisible by 3. If no such group exists, then we are done. If there is such a subgroup, then we can form 'supergroups' in the above manner.
