2
$\begingroup$

This is exercise $1.$ from Section 7: Groups and Homomorphisms, Chapter 1: Foundation, textbook Analysis I by Herbert Amann and Joachim Escher.

Let $N$ be a subgroup of a finite group $(G,\odot)$. Show that $|G| = |N| \cdot |G/N|$ where $G/N$ is the set of left cosets of $G$ modulo $N$.

Does my attempt look fine or contain gaps/flaws? Thank you for your help!


My attempt:

Clearly, $g \odot m \neq g \odot n \iff m \neq n$. Thus $|g \odot N| = |N|$ for all $g \in G$.

We have $G/N \ni [g]=g \odot N$ for all $g \in G$.

Since $G/N$ is a partition of $G$, $|G| = \sum_{[g] \in G/N}|[g]| = \sum_{[g] \in G/N} |N| = |G/N| \cdot |N|$. This completes the proof.

$\endgroup$
  • $\begingroup$ I think this is a duplicate of this. Not using my dupehammer, for I haven't had my morning coffee, yet. $\endgroup$ – Jyrki Lahtonen Mar 18 at 4:57
  • 1
    $\begingroup$ I wouldn't encourage using a dupehammer for someone who wants to verify that their work is correct. $\endgroup$ – Robert Shore Mar 18 at 5:30
1
$\begingroup$

Your proof isn't correct, or at least isn't complete, because you haven't proved that the left cosets form a partition.

Assume $g_1N \cap g_2N =S \neq \emptyset$. Then we need to prove that in fact $g_1N=g_2N.$

Choose $h \in S.$ Then $\exists n_1, n_2 \in N \text{ with }g_1n_1=g_2n_2$ so $g_1n_1n_2^{-1}=g_2 \Rightarrow \forall n \in N~g_2n=g_1n_1n_2^{-1}n \in g_1N \Rightarrow g_2N \subseteq g_1N$. Reverse the roles of $g_1$ and $g_2$ to see also that $g_1N \subseteq g_2N$ and equality follows.

Now we know that the left cosets form a partition of $G$. I'll leave it to you to show that each coset has size $|N|$ and then the result follows.

$\endgroup$
  • 2
    $\begingroup$ I think there is a typo. It should be "...each coset has size $|N|$..." $\endgroup$ – MadnessFor MATH Mar 18 at 5:34
  • 1
    $\begingroup$ Thank you. I've fixed it. $\endgroup$ – Robert Shore Mar 18 at 6:51

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.