# The inequality between indexes of two subgroups of the group

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:

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

• Thanks! Take a look at my edit – ZFR Oct 4 '18 at 20:37
• 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 $|A|\leq |B|$. – 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 $$[G:K] = [G:H][H:K]$$.

• 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. – Mark Oct 4 '18 at 20:43
• @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. – Dietrich Burde Oct 4 '18 at 20:44