I saw the following exercise just now; it should have something to do with group actions:

Let $K\leq A_5$. Show that $|A_5/K|> 4$.

I discovered I don't really understand what this means. First of all, $A_5$ is simple, so $K$ is not normal unless $K=A_5$ or $K=1$. So the quotient set $A_5/K$ is NOT a group. I guess that $|A_5/K|=|A_5: K|=|A_5|/|K|$, but this doesn't seem to be right. Also, when $A_5=K$, $|A_5|/|K|=1$, so this is impossible.

What does this question mean? I am pretty shocked to be unable to understand it.

  • $\begingroup$ The question doesn't seem to make sense to me. Where did you come across it? $\endgroup$ – Math1000 Dec 3 at 0:45
  • $\begingroup$ @Math1000 I saw it here. $\endgroup$ – Jethro Dec 3 at 1:16
  • $\begingroup$ $K$ need not be a normal subgroup. This might be asking about the cardinality of the set of (left) cosets. That set isn't a group, but there is a natural $A_5$-action on it. $\endgroup$ – Ethan Bolker Dec 3 at 1:18
  • $\begingroup$ Yes...but what if $K=A_5$? Then it is not true... $\endgroup$ – Jethro Dec 3 at 1:19
  • $\begingroup$ Perhaps "proper subgroup" is missing by accident. I'm just guessing. There is no subgroup of index $2$. All you have to eliminate next is $3$. $\endgroup$ – Ethan Bolker Dec 3 at 1:20

$G/H$ is standard notation for the set of left cosets of $H$ in $G$. See https://en.wikipedia.org/wiki/Coset#Notation for example. In particular, $|G/H|=|G:H|$, as you surmised.

(When $H$ happens to be normal, then this notation is also used for the natural group on this set, as you know.)

With the notation question out of the way, indeed you need $K<A_5$ for the question to be correct.


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