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I think that isomorphic groups should have the same number of Sylow $p$-groups, but I am not sure why, I am a little stuck on this, I really don't know where to even begin, or if this is even true (sorry I am very new to group theory). I was not able to find anything online. Any help/thoughts?

Thank you!

Nicky Hekster
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Samael Manasseh
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    Isomorphic groups share all algebraic properties, including number of Sylow $p$-subgroups. That's because if $f:G\to H$ is an isomorphism then $K\subseteq G$ is Sylow $p$-subgroup if and only if $f(K)$ is a Sylow $p$-subgroup of $H$. – freakish Mar 29 '22 at 15:30
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    When two groups are isomorphic they are essentially the same group, with different names for the elements. So every question you can ask that uses only group theoretic terms will have the same answer for each group. See https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker Mar 29 '22 at 15:36
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    Isomorphisms of any object are merely a relabeling of the elements. This idea extends beyond algebra into other areas like graph theory too where the relabeling of elements doesn't change the structure you're studying. – CyclotomicField Mar 29 '22 at 15:38

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Yes.

For any isomorphism $\varphi:G\to H$ of groups, we have $K\le G$ if and only if $\varphi(K)\le H$, where $\varphi(K)=\{\varphi(k)\mid k\in K\}$ is the $\varphi$-image of $K$. Therefore, $\varphi$ is a one-to-one correspondence between the subgroups of $G$ with those of $H$.

Moreover, $|G|=|H|$. Isomorphisms preserve the orders of subgroups too, which is to say that $|K|=|\varphi(K)|$ for all $K\le G$.

Can you conclude from here?

Hint: Show that the restriction $$\begin{align}\varphi|_K: K&\to \varphi(K),\\ k&\mapsto \varphi(k)\end{align}$$ of $\varphi$ to $K$ is an isomorphism of $K$ and $\varphi(K)$

Shaun
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    Just because $\varphi$ induces a bijection on subgroups doesn't imply that it maps Sylow $p$-subgroups to Sylow $p$-subgroups, which is essential. It does of course, but this has to be said. – freakish Mar 29 '22 at 15:38
  • Thank you, @freakish; I have edited my answer accordingly. – Shaun Mar 29 '22 at 15:40