I know this question has been posted previously here and here, but they only help with the basic structure of the proof. I feel pretty comfortable up until the part where I start using the 1st Isomorphic Theorem. I just don't see how this how the preimage and this tie together. I know this is the approach to solving it because of the examples in my textbook. I just don't understand why.
Any help would be much appreciated. Thank you!
Below is my attempt and the question I am trying to answer.
Dummit and Foote - Abstract Algebra 3rd Edition
Use Cauchy's theorem and induction to show that a finite abelian group has a subgroup of order $n$ for each positive divisor $n$ of its order. (in the process of solving this problem, you will need to answer the first question of problem 1 on page 84)
Proof: Let $G$ be a finite abelian group. Then $G$ is nonempty. Let $|G|=m$ such that $m\in \mathbb{Z}>0$. Let $n|m$ such that $n\in \mathbb{Z}>0$. By definition of a divisor, $m=sn$ such that $s\in \mathbb{Z}>0$. Therefore $|G|=m\geq n$. We want to show that for each $n$ that divides $m$ there is a subgroup in $G$ of order $n$.
We will prove this by using induction on the order of $G$. Let $|G|=1$ be the base case. This means $G=\{1\}$. Then $n=1$ because 1 is the only divisor of 1. Notice $G\leqslant G$, and we are done. Assume true for all finite abelian groups with order less than $|G|$. Now we will consider the cases when $|G|$ is prime and when it is composite.
Suppose $|G|=m$ such that $m$ is a prime number. Then $n=p$ or $n=1$ because these are the only divisors of a prime number. If $n=1$, then $\{1\}\leqslant G$. If $n=p$, then $G\leqslant G$. Thus this property holds when the order of $G$ is prime.
Suppose $|G|=m$ such that $m$ is not prime and $m\in\mathbb{Z}>0$. Then $G$ is composite. By the Fundamental Theorem of Arithmetic, $m$ can be factored uniquely into the product of primes. This means there are distinct primes $p_1,p_2,\ldots,p_s$ and positive integers $\alpha_1,\alpha_2,\ldots,\alpha_s$ such that $$ m=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_s^{\alpha_s} $$ Let $p_i|m$. Since $G$ is a finite group and $p_i|m$, then $G$ has an element, say $g$, of order $p_i$ by the Cauchy Theorem. Let $H=\langle g\rangle=\{g^a|a\in \mathbb{Z}\} $. Then $|H|=|g|=p_i$ by Proposition 2 (pg. 55). Now we need to show that $H$ is a subgroup of $G$. Since $g\in\langle g\rangle=H$, $H$ is nonempty. Let $g^x,g^y\in H$ such that $x,y\in \mathbb{Z}$. Then $$ g^x(g^y)^{-1} = g^{x-y}\in\langle g \rangle=H $$ Therefore $H$ is a subgroup $G$. Since $H\leqslant G$ and $G$ is finite, then $H$ is a finite subgroup. Also notice that $H$ is cyclic by definition because it can be generated by a single element. This means $H=\langle g\rangle$ is a finite cyclic subgroup of $G$.
Since $G$ is an abelian group, any subgroup of $G$ is normal (pg. 84). Then $H$ is normal in $G$. Consider the quotient group $G/ H$. By Lagrange, $$ \left| G/ H \right| = \frac{|G|}{|H|} = \frac{m}{p_i}. $$ Therefore it is clear that $|H|$ has one less prime divisor than $|G|$. Since a quotient group reflects the structure of $G$ when $H$ is normal (pg. 82), $G/H$ is a finite abelian group with order less than $|G|$. Notice $n|m$ implies $\frac{n}{p_i}\left|\right. \frac{m}{p_i}$. Then by the induction hypothesis, $K\leqslant G/H$ where $|K|=\frac{n}{p_i}$ because $\frac{n}{p_i}\left|\right. \frac{m}{p_i}$.
Now consider the quotient homomorphism $\varphi: G\rightarrow G/H$. Recall $K\leqslant G/H$. We want to show that $\varphi^{-1}(K)\leqslant G$. Let $x,y\in\varphi^{-1}(K)$. Then $\varphi(x),\varphi(y)\in K$. Since $K\leqslant G/H$, then $\varphi(xy^{-1})\in K$ by the one-step subgroup test. This implies $xy^{-1}\in \varphi^{-1}(K)$. Therefore by the one-step subgroup test, $\varphi^{-1}(K)\leqslant G$. This means the preimage or pullback of a subgroup under a homomorphism is a subgroup
By the 1st Isomorphic Theorem, $G/\ker(\varphi)\cong \varphi(G)$.
