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, $G/\ker(\varphi)\cong \varphi(G)$.


Maybe the Third Isomorphism Theorem is more appropriate here than the First Isomorphism Theorem. It says that for a group $G$ and normal subgroups $H \leq N \leq G$, we have

$$ (G/H)/(N/H) \cong G/N $$

In your case, $N/H$ is what you call $K$, the subgroup of $G/H$ of size $n/p_i$. Then $N$ will be what you call $\varphi^{-1}(K)$, the pre-image in $G$, where $\varphi$ is the quotient map onto $G/H$. Take the statement from the theorem, and write the orders/cardinalities of the groups to get the size of $N = \varphi^{-1}(K)$.

Since, as you mention, $|G/H| = m/p_i$, and $|K| = |N/H| = n/p_i$, this gives

$$ \left( \frac{m}{p_i} \right) / \left( \frac{n}{p_i} \right) = \frac{m}{n} = \frac{m}{|\varphi^{-1}(K)|} $$

This shows that $|\varphi^{-1}(K)| = n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.