An exercise from an "Algebra: Chapter 0" textbook: uniqueness of the decomposition of a $p$-Sylow subgroup

Question

This is an exercise of "Algebra: Chapter 0" by Paolo Aluffi:

Prove that the decomposition of a finite abelian group $G$ as a direct sum of cyclic $p$-groups is unique. (Hint: The prime factorization of $|G|$ determined the primes, so it suffices to show that if

$$\frac{\mathbb{Z}}{p^{r_1}\mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m}\mathbb{Z}} \cong \frac{\mathbb{Z}}{p^{s_1}\mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{s_n}\mathbb{Z}}$$

with $r_1 \geq ... \geq r_m$ and $s_1 \geq ... \geq s_n$, then $m = n$ and $r_i = s_i$ for all $i$. Do this by induction, by considering the group $pG$ obtained as the image of the homomorphism $\phi: G \to G$ defined by $\phi(g) = pg$.)

Maybe, I should clarify. We know that any finite group $G, \ |G| = p^{r_1}_1...p^{r_k}_k$ decomposes as the direct sum of it's ($p_i$-)Sylow subgroups(we can donte the $p_i$-Sylow subgroup(which is unique since $G$ is abelian) by $P_i$.

And each $P_i$ decomposes as a direct sum of cyclic $p_i$-groups, namely

$P_i \cong \frac{\mathbb{Z}}{p_1^{r_1}\mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p_1^{r_m}\mathbb{Z}}$

Since each $P_i$ is uniquely determined by $|G|$, all we need to prove that the decomposition of $P_i$ into cyclic $p$-groups in unique.

Any hints on how $\phi$ and $pG$ may play the role? First Isomorphism Theorem tells us $pG \cong \frac{G}{ \{0 \} \cup \{ g \in G \ | \ |g| = p \}}$

Any help would be appreciated. Note that this isn't simple "prove uniqueness of the decomposition question", here I'm interested in getting hints on the aaproach proposed by P.Aluffi.

Answer 1

I think it is resolved. First of all, a somewhat of a "lemma type".

Lemma 1. let $G_1$ and $G_2$ be abelian groups. $G_1 \cong G_2 \Rightarrow pG_1 \cong pG_2$

Proof: Let $\phi: G_1 \to G_2$ be a group isomorphism. Define $\phi_p: pG_1 \to pG_2$ by setting $\phi_p(g_1^p) = \phi(g_1)^p$. It is a restriction of a function $\phi$.

To prove it is well-defined we need to show that $\forall x \in pG_1 \ \ \phi(x) \in pG_2$

$\phi_p(pg_1) = p\phi(g_1) \in pG_2$. So it is well-defined.

It inherits the properties of being a homomorphism and injectivity from $\phi$. All that is left is to prove it is surjective.

Let $pg_2 \in pG_2$. Since $\phi$ is surjective, $g_2 = \phi(g_1)$ for some $g_1 \in G$. Then $pg_2 = p\phi(g_1) = \phi(pg_1)$.

So $\phi_p$ is an isomorphism of groups.

Now, the proof of Lemma 1 is finished and we can move on to Lemma 2.

Lemma 2. $p(\frac{\mathbb{Z}}{p^{r_1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m} \mathbb{Z}}) \cong \frac{\mathbb{Z}}{p^{r_1 - 1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m - 1} \mathbb{Z}}$. For a proof, visit https://math.stackexchange.com/a/1921016/229776

Now, the proof of the main statement. We do the proof by induction. Remember that $r_1 \geq ... \geq r_m > 0$ and $s_1 \geq ... \geq s_n > 0$. We induct on $k = r_1 + ... + r_m = s_1 + ... + s_n$. If $k = 1$, then $m = n = 1$ and $r_1 = s_1 = 1$(and both groups are the group $\mathbb{Z}/ p \mathbb{Z}$ ).

Now we assume the inductive hypothesis and go for an arbitrary $k$.

If $\frac{\mathbb{Z}}{p^{r_1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m} \mathbb{Z}} \cong \frac{\mathbb{Z}}{p^{s_1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{s_n} \mathbb{Z}}$, then by Lemma 1 and Lemma 2 we have

$\frac{\mathbb{Z}}{p^{r_1-1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m-1} \mathbb{Z}} \cong \frac{\mathbb{Z}}{p^{s_1 - 1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{s_n - 1} \mathbb{Z}}$.

We know that $r_1 + ... + r_m - m = s_1 + ... + s_n - n = 1 < k$

Now seems like a good time to use the induction hypothesis, but first, we need to consider a special case.

If $r_1 = 1$, then so is $r_i$ for $1 < i \geq m$. Then after multiplying both groups by $p$(see Lemma 2) we get(see Lemma 1):

$\{0\} \cong \bigoplus_{j=1}^{m} \mathbb{Z}/ p^{r_j-1} \mathbb{Z}$

So we conclude that $s_1 = 1$ also.

So we have

$\underbrace{ \mathbb{Z}/p \mathbb{Z} \oplus \ldots \oplus \mathbb{Z}/p \mathbb{Z}}_{n \text{ times}} \cong \underbrace{ \mathbb{Z}/p \mathbb{Z} \oplus \ldots \oplus \mathbb{Z}/p \mathbb{Z}}_{m \text{ times}}$

From order/cardinality considerations, $p^m = p^n,$ so $m = n$.

The same arguement works for the case $s_1 = 1$.

Now, let $s_1, r_1 > 1$.

Let $t$ be the largest number $i$ such that $r_i > 1$ and let $f$ be the largest number such that $s_j > 1$.

Then we have(after multiplying both groups by $p$):

$\bigoplus_{i=1}^{t} \mathbb{Z}/ p^{r_i} \mathbb{Z} \cong \bigoplus_{j=1}^{f} \mathbb{Z}/ p^{s_j} \mathbb{Z}$

By inductive hypothesis we have $t = f$ and $r_i - 1 = s_i - 1 \Leftrightarrow r_i = s_i$ for all $i \in \{1, ..., f \}$.

Now, it means that

$\frac{\mathbb{Z}}{p^{r_1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_t} \mathbb{Z}} \oplus \underbrace{ \frac{\mathbb{Z}}{p \mathbb{Z}} \oplus \ldots \oplus \frac{\mathbb{Z}}{p \mathbb{Z}}}_{d_1 \text{ times}} \cong \frac{\mathbb{Z}}{p^{r_1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_t} \mathbb{Z}} \oplus \underbrace{ \frac{\mathbb{Z}}{p \mathbb{Z}} \oplus \ldots \oplus \frac{\mathbb{Z}}{p \mathbb{Z}}}_{d_2 \text{ times}}$

From order considerations we get $d_1 = d_2$.

Answer 2

Since you mentioned Sylow subgroups, we could use Cauchy's theorem, saying that every group of order $n$ has, for every prime divisor $p$ of $n$, an element of order $p$. Then we can use induction to show that any finite abelian group is a direct sum of cyclic subgroups of prime-power order in a unique way. A very detailed proof is given in Sullivan's notes on page $2$.