Suppose that $G$ is a group and $H$ and $K$ are its subgroups with $K\subset H$. Prove that $[G:H]\leq [G:K]$.

Intuitively i know that this is true. But I am trying to prove it in rigorous way: let's denote the set of all left cosets of $K$ in $G$ by $G/K$ and by $G/H$ the set of all left cosets of $H$ in $G$.

Consider the mapping $\varphi:G/K\to G/H$ defined by $\varphi(gK)=gH$ for any $g\in G$. It's easy to check that this map is well-defined and surjective. How to conclude rigorously the need inequality?

Would be very grateful for help!

EDIT: If $f:X\to Y$ the mapping between two sets and $f$ - surjective. Then two cases are possible:

1) If $Y$ is infinite then $X$ is also infinite.

2) If $Y$ is finite then $|X|\geq |Y|$.

In both cases we get that $|X|\geq |Y|$.

Is my reasoning correct?


Well, that's it. If there is a surjective function from set $A$ to set $B$ then the cardinality of set $A$ must be bigger or equal to the cardinality of $B$. So here:

$[G:K]=|G/K|\geq |G/H|=[G:H]$

  • $\begingroup$ Thanks! Take a look at my edit $\endgroup$ – ZFR Oct 4 '18 at 20:37
  • $\begingroup$ Well, note that the infinite cardinality is not unique. For example $|\mathbb{R}|>|\mathbb{N}|$. So it's not right just to say that if two sets are infinite then their cardinalities are equal. But if there is a surjective function from $A$ to $B$ then by one of the definitions of cardinality we say $|A|\leq |B|$. $\endgroup$ – Mark Oct 4 '18 at 20:41

Because of $[G:K]=[G:H][H:K]$ and $[H:K]\ge 1$ we have $[G:K]\ge [G:H]$.

Reference for the degree theorem: If $K \leq H \leq G$, show that $[G:K] = [G:H][H:K]$.

  • 1
    $\begingroup$ This equation is not trivial and has to be proved. It is easy to do for finite groups (Lagrange's theorem) but for infinite groups it actually requires a bit of work. $\endgroup$ – Mark Oct 4 '18 at 20:43
  • $\begingroup$ @Mark Yes, you are right, but we already have done it here (duplicate...) Of course, an easier proof is also useful, but the degree theorem anyway is needed lateron. $\endgroup$ – Dietrich Burde Oct 4 '18 at 20:44
  • $\begingroup$ Yes, I just wrote my comment before you edited your answer. I mean OP might not know about this equation. But now you added a link, so it's ok. $\endgroup$ – Mark Oct 4 '18 at 20:46

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