2
$\begingroup$

One of the isomorphism theorems states $(HN) \ / \ N \cong H \ / \ (H\cap N)$.

I am confused about the first part $(HN) \ / \ N$, and whether it is equivalent to $H \ / \ N$. Every element in $HN$ is $hn$ for some $h \in H, n \in N$. So then every coset in $(HN) \ / \ N$ can be expressed as $\{hnN \text{ for some } h \in H, n \in N\}$, but this is just the same as $\{hN \text{ for some } h \in H\}$, so why isn't the first isomorphism theorem just

$H \ / \ N \cong H \ / \ (H\cap N)$

$\endgroup$
4
  • $\begingroup$ Not sure this is the first isomorphism theorem. $\endgroup$
    – Kan't
    Apr 18 at 20:58
  • 3
    $\begingroup$ @Kan't It is in some literature. $\endgroup$
    – azif00
    Apr 18 at 21:01
  • $\begingroup$ As far as I could survey so far, I didn't find any, @azif00. Could you share/mention a such one, please? $\endgroup$
    – Kan't
    Apr 20 at 8:09
  • $\begingroup$ Abstract Algebra by Blair and Beachy $\endgroup$ Apr 21 at 22:23

1 Answer 1

5
$\begingroup$

The notation $A/B$ only makes sense (as a group) when $B$ is a normal subgroup of $A$. In general, we do not require that $H$ contains $N$ for the theorem holds.

However, the equality $HN/N = \{hN : h \in H\}$ is correct.

$\endgroup$
2
  • $\begingroup$ I added "(as a group)" because some people study the set of left cosets, denoted by $A/B$, and the notation makes sense as a group when it coincides with right cosets, denoted $B\setminus A$, so when $B\unlhd A$. $\endgroup$
    – Shaun
    Apr 18 at 21:11
  • 1
    $\begingroup$ @Shaun Yes, I agree. $\endgroup$
    – azif00
    Apr 18 at 21:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .