every Abelian group is a converse lagrange theorem group

Let $G$ be a finite abelian group, then $G$ has a subgroup of order $n$ if and only if $n\mid G$.

Proof: by Lagrange if $H\leq G$ then $|H|$ divides $|G|$ so this proves one of the implications.

We prove the other implication by strong induction on the order of $G$, for primes this is just Cauchy, so we can take these like base cases. Now let $n$ be a proper divisor of $|G|$, take $p$ a prime in the factorization of $|G|$ (instead of $|G|$ is should be $n$, I had made a mistake when typing the first time), then by cauchy there is an element $x$ of $G$ of order $p$, since $G$ is abelian $\langle x\rangle$ is normal in $G$ so we can look at the quotient group $G/\langle x\rangle$, this group has order $\frac{|G|}{p}$, now take a prime $p_1$ in the factorization of $\frac{|G|}{n}$. By induction the quotient group has a subgroup $E$ of order $G/pp_1$ now look at the preimage of $E$ under the natural homomorphism from $G$ to $G/\langle x\rangle$, it is easy to see this is a subgroup of $G$, we call it $I$, consider the restriction of the natural projection to $I$, this is a surjective homomorphism from $I$ to $E$ with kernel $\langle x\rangle$ so by the first isomorphism theorem $I/x\cong E$ so $\frac{|I|}{|\langle x\rangle|}=|E|\implies |I|=\frac{|G|}{pp_1}{p}=\frac{|G|}{p_1}$ Notice $\frac{|G|}{p_1}$ is a multiple of $n$ since $p_1$ was in the fatorization of $\frac{|G|}{n}$. So by induction $I$ has a subgroup of order $n$ which is also a subgroup of $G$.

Solution 1:

Your proof is definitely on the right track. The first part looks good. For the second part, there are just a few small issues.

• If you are using strong induction on $|G|$, then your base is for the $|G|=1$. You still need Cauchy's theorem later on, to mod out by a group of order $p$.

• Your argument breaks down in one case: suppose that $G=qn$, for $q$ a prime with $\gcd(n,q)=1$ (note this implies that $q^2$ does not divide $|G|$). Suppose we take $p=q$ in your argument. At the next step, we also must choose $p_1=q$, since this is the only prime dividing $\frac{|G|}{n}$. However, $G/\langle x \rangle$ contains no subgroup of order $\frac{|G|}{pp_1}$, since $pp_1=q^2$ does not divide the order of $G$. You should be able to deal with this by being a little bit more specific about how $p$ is chosen.

Alternately, you could explicitly require that $p$ be a divisor of $n$ in your argument. Then a subgroup of order $n$ can be constructed as the preimage of a subgroup under the natural projection.

That is, given a group $G$ with $|G|>1$, let $n$ be a divisor of $|G|$. Write $n=pk$ for $p$ a prime, and let $H$ be a subgroup of $G$ of order $p$ (appealing to Cauchy's theorem). By strong induction on the size of the group, $G/H$ has a subgroup $K$ of order $k$... Do you see how to finish?