# Let $N$ be a subgroup of a finite group $G$. Show that $|G| = |N| \cdot |G/N|$

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.

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

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