1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $[G:K] = [G:H][H:K]$.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.