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If $H,K$ are subgroups of finite group $G,$ then $|G:(H \cap K)|\leq |G:H||G:K|$

May I know if my proof is correct? Thank you v. much.

Proof:

$$|G:(H \cap K)| \leq |G:H||G:K| \Longleftrightarrow \frac{|G|}{|H \cap K|}\frac{|H|}{|G|} = \frac{|H|}{|H \cap K|}\leq \frac{|G|}{|K|}$$

So it suffices to prove: $|H/(H\cap K)| \leq |G/K|,$ where $G$ is not necessarily finite. Given $h \in H,$ let $$ h(H\cap K) \mapsto \ hK$$

The mapping is well-defined:

$$h'(H \cap K) = h(H \cap K) \implies \exists m \in H \cap K: h'=hm \implies h'K = hmK= hK$$

The mapping is injective:

$$hK = h'K \implies \exists k \in K: h=h'k \implies k \in H \cap K \implies h(H\cap K) = h'(H\cap K)$$

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    $\begingroup$ I would recommend getting used to writing proof with much less logic notation. I had it too and it's a bad habit. Additional i found that writing something in clear english demands a higher level of understanding than writing it logically so it's another good way of testing your understanding. $\endgroup$ – Saal Hardali Nov 26 '13 at 8:17
  • $\begingroup$ Less logic notation than the used in the OP?? I don't agree: usual, standard logical symbols (e.g. $\;\forall\,,\,\exists\;,\;\implies\;,\;\iff\;$, etc.) make things much clearer and shorter, something most of us mathematicians cannot have too much of. $\endgroup$ – DonAntonio Nov 26 '13 at 13:56
  • $\begingroup$ Although many mathematicians do it anyway, it is technically agrammatical to use $\Rightarrow$ as a synonym for "thus." What it really means is "implies." If you really want to write "thus" as a symbol, you should use ∴ instead. Personally I'm a big fan of "so" for long calculations, e.g. "Let $\text{P}$ be true, then $\text{Q}$ is true so $\text{R}$ is true so $\text{S}$ is true so $\text{T}$ is true..." $\endgroup$ – Alexander Gruber Nov 26 '13 at 21:40
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$\renewcommand{\phi}{\varphi}$I think it is preferable to prove first a result of general interest, that is $$ \lvert H K : K \rvert = \lvert H : H \cap K \rvert. $$ Here $HK$ is not necessarily a subgroup, but it is definitely a union of right cosets $h K$ of $K$, for $h \in H$.

This I believe you have implicitly proved, anyway it goes as follows. Compose the map $H \to H K$ given by $h \mapsto h \cdot 1$ with the map $HK \to HK/K$ given by $g \mapsto gK$ to get the map $\phi: H \mapsto HK/K$ given by $h \mapsto h K$. This is clearly onto, and for $a, b \in H$ one has \begin{align} \phi(a) = \phi(b) \text{ iff }& a K = b K \\\text{iff } & a^{-1} b \in H \cap K \\\text{iff }& a (H \cap K) = b (H \cap K). \end{align} Thus $\phi$ induces a bijection $H/H \cap K \to HK/K$.

Now argue as you did $$ \lvert G : H \cap K \rvert = \lvert G : H \rvert \cdot \lvert H : H \cap K \rvert = \lvert G : H \rvert \cdot \lvert HK : K \rvert \le \lvert G : H \rvert \cdot \lvert G : K \rvert. $$

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A simpler solution is to prove that the map

$ \qquad G/(H\cap K) \to G/H \times G/K$ given by $ g(H\cap K) \mapsto (gH, gK) $

is well defined and injective.

(The quotients here are sets of cosets.)

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