I'm currently studying I. N. Herstein's Topics in Algebra, which provides three proofs of Sylow's Theorem. I'm working through the second proof (pp. 94-95) at the moment. I can follow every step in the proof except one short section at the end.

The proof works by induction on $n$, the order of a group $G$. The goal is to show that if $p^k$ divides $n$, then $G$ has a subgroup of order $p^k$ (where $p$ is prime and $k\geq1$).

After some steps, we consider the case when $p$ divides $|Z(G)|$, the order of the center of the $G$. Since $p$ is prime, Cauchy's Theorem guarantees that there is an element $b\in \text Z(G)$, with order $p$. Let $\langle b\rangle$ denote the cyclic subgroup generated by $b$. Since $b\in \text Z(G)$, we know that $\langle b\rangle$ is normal in $G$. Hence the quotient group $G/\langle b\rangle$ is defined. By Lagrange's Theorem, $|G/\langle b\rangle|=n/p$.

By assumption $p^k$ divides $n$, so $p^{k-1}$ divides $n/p$. Of course $n/p<n$, so by the induction hypothesis, $G/\langle b\rangle$ has a subgroup of order $p^{k-1}$. Call this subgroup $\overline P$. Now suppose $\phi:G\to G/\langle b\rangle$ is the canonical quotient map. Then the inverse image $P=\phi^{-1}(\overline P)$ is a subgroup of $G$. Here is the part where I'm confused: now Herstein claims that $\overline P\approx P/\langle b\rangle$.

For some reason I'm getting that $\overline P=P/\langle b\rangle$, i.e. the two groups are equal, not just isomorphic. Indeed, if a coset $\langle b\rangle x\in\overline P$, then $x\in P$, so $\langle b\rangle x\in P/\langle b\rangle$. Conversely, if $\langle b\rangle x\in P/\langle b\rangle$, then $x\in P$, so $\langle b\rangle x\in\overline P$.

Is my reasoning correct? Sorry for the lengthy question, I hope that too much information is better than too little. I uploaded a picture of the proof from the book here: https://imgur.com/a/TmQfLT5

(Regardless, I understand that $|\overline P|=|P/\langle b\rangle|$, from which we can apply Lagrange's Theorem to conclude that $|P|=p^k$, which is exactly the subgroup we want, completing the proof.)

  • $\begingroup$ Since $\phi$ is surjective, we have $\phi(\phi^{-1}(\overline P) = \overline P$. The left hand side is $\phi(P) = P / \langle b \rangle$ (for an arbitrary subgroup $P \leq G$ we would have $\phi(P) = P\langle b \rangle / \langle b \rangle$, but this equals $P/ \langle b \rangle$ if $\langle b \rangle \leq P$, which is indeed the case for your particular $P$ by its definition). So you are correct that $\overline P = P / \langle b \rangle$. $\endgroup$ – Bungo May 8 at 2:34

They are indeed equal, if you are using the most common definition of a quotient group. This is just an instance of a writer not wanting to think too hard and using the "safe" $\cong$ symbol instead of $=$.

To be fair, two groups being isomorphic is exactly as powerful as them being equal (as far as group theory is concerned) so this is probably an "error" that is worth making for the time spent not worrying about it.

EDIT: Rereading your post, I want to emphasize my last point -- groups being isomorphic may as well be equal. You have a "proof" of the fact that $\overline P = P/\langle b\rangle$, but in some sense you're just restating the definition. Two groups being "equal" is such a powerful condition (if you could even call it that) that really the only time you're allowed to say this is exactly when you've previously declared them to be equal!

As an example, consider the group $\mathbb Z/6 \mathbb Z$ defined the usual way (as a quotient of $\mathbb Z$) and a group $G$ that I'll define as follows: the underlying set is $\{0, 1, 2, 3, 4, 5\}$, and the group operation is addition modulo $6$. It is beyond clear that $G \cong \mathbb Z/6\mathbb Z$; in fact when most people say $\mathbb Z/6\mathbb Z$ they're really thinking about $G$. Yet they aren't equal -- clearly the underlying set is different. In short, it's pointless to worry about group equality.

| cite | improve this answer | |

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.