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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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.

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

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