If $H$ is a subgroup of a finite abelian group $G$, then $G$ has a subgroup that is isomorphic to $G/H$.

I know Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$? has two answers. I don't understand how the first answer works and I have doubt about that answer. The second answer uses character theory which I don't intend to use since I am practicing for my qualifying exam. This problem showed up in our last exam. None of my classmates can solve this problem and the professors in our department just provided us with some hints saying look the homomorphism and do induction which helped a little but didn't lead me to an answer.

Now I list the two approaches I have tried and the reasons why they don't work.

The first approach I tried is the following: Suppose $G$ is finite abelian. Let $|G|=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ and $|H|=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$. By the elementary divisor decomposition, we have $G\cong G_1\times G_2\times\cdots\times G_k$ and $H\cong H_1\times H_2\times\cdots\times H_k$ where $G_i$ is a Sylow subgroup of $G$ and $H_i$ is a Sylow subgroup of $H$ for all $i=1,2,...,k$. Hence $G/H\cong(G_1/H_1)\times(G_2/H_2)\times\cdots\times(G_k/H_k)$. So it suffices to show the result when $G$ is a $p$-group. Suppose $|G|=p^n$.

Now by the invariant factor decomposition, $G\cong\mathbb{Z}_{p^{a_1}}\times\mathbb{Z}_{p^{a_2}}\times\cdots\times\mathbb{Z}_{p^{a_t}}$ with $a_1\geq a_2\geq\cdots\geq a_t$ and $a_1+a_2+\cdots+a_t=n$. For $i=1,...,t$, let $e_i=(...,0,1,0,...)$. Then $G/H=\left<e_1H,...,e_tH\right>$. WLOG, let $\{e_1H,e_2H,...,e_sH\}$, $s\leq t$, be a smallest set of generators for $G/H$.

Claim: $G/H\cong\left<e_1H\right>\times\cdots\times\left<e_sH\right>$.

So $G/H\cong\mathbb{Z}_{p^{b_1}}\times\cdots\times\mathbb{Z}_{p^{b_s}}$ with $b_i\leq a_i$ for all $i=1,...,s$. (This is due to the canonical projection $G\mapsto G/H$ and thus $|e_iH|\mid |e_i|$ for all $i=1,...,s$.) Therefore, $G/H$ is isomorphic to a subgroup of $G$.

This looks promising, but the claim is false. A counterexample is this: Let $G=\mathbb{Z}_4\times\mathbb{Z}_4$. Then $G/\left<(2,2)\right>\cong\mathbb{Z}_4\times\mathbb{Z}_2$, but $(0,1)+\left<(2,2)\right>$ and $(1,0)+\left<(2,2)\right>$ both have order 4.

The second aprroach I tried is the following: Suppose $G$ is finite abelian and $H\leq G$. Let $|G|=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ and $|H|=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$ where $p_1,...,p_k$ are distinct primes. By the elementary divisor decomposition, we have $G\cong G_1\times G_2\times\cdots\times G_k$ and $H\cong H_1\times H_2\times\cdots\times H_k$ where $G_i$ is a Sylow $p_i$-subgroup of $G$ and $H_i$ is a Sylow $p_i$-subgroup of $H$ for all $i=1,2,...,k$. Since $H_i\unlhd G_i$ for all $i=1,...,k$, $G/H\cong(G_1/H_1)\times(G_2/H_2)\times\cdots\times(G_k/H_k)$. So it suffices to show the result when $G$ is an abelian $p$-group. We proceed by induction.

If $|G|=p$, then $H=1$ or $G$, so $G/H\cong1$ or $G$.

Suppose the result holds for all abelian $p$-groups of order less than $|G|$. Now by the fundamental theorem of finite abelian groups, $G=\mathbb{Z}_{p^{\alpha_1}}\times\cdots\times\mathbb{Z}_{p^{\alpha_n}}=\left<x_1\right>\times\cdots\times\left<x_n\right>$. Consider $\varphi:G\to G$ such that $x\mapsto x^p$. Since $G$ is abelian, $\varphi$ is a group homomorphism with $\ker\varphi=\left<x_1^{p^{\alpha_1-1}}\right>\times\cdots\left<x_n^{p^{\alpha_n-1}}\right>$. By Cauchy's theorem, $H':=\ker\varphi\cap H\neq1$ and it is elementary abelian. So WLOG, $H'\cong\left<x_1^{p^{\alpha_1-1}}\right>\times\cdots\left<x_m^{p^{\alpha_m-1}}\right>$ where $m\leq n$. It follows that $G/H'$ is isomorphic to a subgroup of $G$. By the third isomorphism theorem, we have $G/H\cong(G/H')/(H/H')$. Since $|G/H'|<|G|$, by the induction hypothesis, $(G/H')/(H/H')$ is isomorphic to a subgroup of $G/H'$ and thus it is isomorphic to a subgroup of $G$. Therefore, $G/H$ is isomorphic to a subgroup of $G$.

Now in this proof I assume that if $B$ and $C$ are isomorphic subgroups of a finite abelian group $A$, then $A/B\cong A/C$ which is not true. A counterexample is $A=\mathbb{Z}_4\times\mathbb{Z}_2$, $B=\left<(2,0)\right>$ and $C=\left<(0,1)\right>$. Here $A/B\cong\mathbb{Z}_2\times\mathbb{Z}_2\not\cong\mathbb{Z}_4\cong A/C$.

Now my question is does anyone know how to fix either of the problems in my two attempts above to make it work? Or have a better explanation for the first answer in the post Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$?? This problem have haunted me for months. I would really appreciate the help.

Solution 1:

Since no one answered my question, I did some reading and found out that this is a very well known result. By using the concept of a "basis" of an abelian group, I did the following proof.

Suppose $G$ is a finite abelian group and $H\leq G$. Let $|G|=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ and $|H|=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$ where $p_1,...,p_k$ are distinct primes. By the elementary divisor decomposition, we have $G\cong G_1\oplus G_2\oplus\cdots\oplus G_k$ and $H\cong H_1\oplus H_2\oplus\cdots\oplus H_k$ where $G_i$ is a Sylow $p_i$-subgroup of $G$ and $H_i$ is a Sylow $p_i$-subgroup of $H$ for all $i=1,2,...,k$. Since $H_i\unlhd G_i$ for all $i=1,...,k$, $G/H\cong(G_1/H_1)\oplus(G_2/H_2)\oplus\cdots\oplus(G_k/H_k)$. So it suffices to show the result when $G$ is an abelian $p$-group. We proceed by induction.

Notice it is easier to think of $G$ as an additive group in stead of a direct product.

If $|G|=p$, then $H=1$ or $G$, so $G/H\cong1$ or $G$.

Suppose the result holds for all abelian $p$-groups of order less than $|G|$. Since $G$ is an abelian $p$ group, by the fundamental theorem of finite abelian groups, $G=\left<x_1\right>\oplus\left<x_2\right>\oplus\cdots\oplus\left<x_t\right>$ with $|x_i|=p^{\alpha_i}$ for all $i\in\{1,...,t\}$ and $\alpha_1\geq\alpha_2\geq\cdots\geq\alpha_t\geq1$. Notice that $x_1,x_2,...,x_t$ are linearly independent in the sense that $x_i$ cannot be write as a linear combination of $x_1,...,x_{i-1},x_{i+1},...,x_t$.

Case 1: $H=\left<g\right>$ with $|g|=p$. Since $|g|=p$, $g=m_1p^{\alpha_1-1}x_1+m_2p^{\alpha_2-1}x_2+\cdots+m_tp^{\alpha_t-1}x_t$ with $m_i\in\{0,1,...,p-1\}$ for all $i\in\{1,...,t\}$ (hence, $m_i=0$ or $(m_i,p)=1$). WLOG, assume $m_t\neq0$. Let $x_t'=m_1p^{\alpha_1-\alpha_t}x_1+m_2p^{\alpha_2-\alpha_t}x_2+\cdots+m_{t-1}p^{\alpha_{t-1}-\alpha_t}x_{t-1}+m_tx_t$. Notice that $|x_t'|=p^{\alpha_t}$. Since $\left<x_t\right>\cap(\left<x_1\right>\oplus\cdots\oplus\left<x_{t-1}\right>)=0$, $\left<x_t'\right>\cap\left<x_i\right>=0$ for all $i\in\{0,1,...,t-1\}$ otherwise $x_t'$ would be a linear combination of $x_1,...,x_{t-1}$. Hence, $G=\left<x_1\right>\oplus\left<x_2\right>\oplus\cdots\oplus\left<x_t'\right>$ (This is basically changing the basis of $G$). Since $\left<g\right>\leq\left<x_t'\right>$, $G/H\cong\left<x_1\right>\oplus\left<x_2\right>\oplus\cdots\oplus(\left<x_t'\right>/\left<g\right>)$. Since $\left<x_t'\right>$ is cyclic, $\left<x_t'\right>/\left<g\right>$ is isomorphic to a subgroup of $\left<x_t'\right>$. So $G/H$ is isomorphic to a subgroup of $G$.

Case 2: $|H|>p$. By Cauchy's theorem, there exists $g\in H$ such that $|g|=p$. By Case 1, $G/\left<g\right>$ is isomorphic to a subgroup of $G$. Now by the third isomorphism theorem, $G/H\cong(G/\left<g\right>)/(H/\left<g\right>)$. Since $|G/\left<g\right>|<|G|$, by the induction hypothesis, $(G/\left<g\right>)/(H/\left<g\right>)$ is isomorphic to a subgroup of $G/\left<g\right>$ which is isomorphic to a subgroup of $G$. Hence $G/H$ is isomorphic to a subgroup of $G$.