1. ## Third Iso Theorem

Let G be a group and let H and K be normal subgroups of G with H a subgroup of K. Then $\displaystyle K/H\trianglelefteq G/H$ and $\displaystyle (G/H)/(K/H)\cong G/K$

How do I show:
$\displaystyle K/H\trianglelefteq G/H$

2. ## Re: Third Iso Theorem

most of the work here is showing that K/H is a subgroup of G/H. normality quickly follows.

(H is clearly normal in K, since it's normal in a larger group G).

so consider Hk, Hk' in K/H. is Hkk'^-1 is K/H?

what would it mean for K/H to be normal in G/H?

we would need (Hg)(K/H)(Hg)^-1 to be in K/H.

an element of (Hg)(K/H)(Hg)^-1 is the coset product: (Hg)(Hk)(Hg)^-1 = Hgkg^-1.

but K is normal in G, so....

3. ## Re: Third Iso Theorem

define the map $\displaystyle \phi: G/H \longrightarrow G/K$ by $\displaystyle \phi(gH)=gK,$ which is well-defined because $\displaystyle H \subseteq K$. clearly $\displaystyle \phi$ is a group homomorphism and $\displaystyle \ker \phi = K/H$. thus, by the first isomorphism theorem, $\displaystyle K/H$ is a normal subgroup of $\displaystyle G/H$ and $\displaystyle (G/H)/(K/H) \cong G/K$.

4. ## Re: Third Iso Theorem

Originally Posted by NonCommAlg
define the map $\displaystyle \phi: G/H \longrightarrow G/K$ by $\displaystyle \phi(gH)=gK,$ which is well-defined because $\displaystyle H \subseteq K$. clearly $\displaystyle \phi$ is a group homomorphism and $\displaystyle \ker \phi = K/H$. thus, by the first isomorphism theorem, $\displaystyle K/H$ is a normal subgroup of $\displaystyle G/H$ and $\displaystyle (G/H)/(K/H) \cong G/K$.
is it just me, or does there seem to be a spooky link between this question, and this one:

http://www.mathhelpforum.com/math-he...es-191667.html ?

5. ## Re: Third Iso Theorem

Originally Posted by Deveno
is it just me, or does there seem to be a spooky link between this question, and this one:

http://www.mathhelpforum.com/math-he...es-191667.html ?
not quite! let $\displaystyle A=\{v_i + N : \ i \in I\}$ be a basis for $\displaystyle V/N$ and suppose that $\displaystyle \sum_{i \in J} c_i(v_i + M) = 0$ for some finite set $\displaystyle J \subseteq I$ and scalars $\displaystyle c_i.$ then $\displaystyle \sum_{i \in J} c_iv_i \in M \subseteq N$ and so $\displaystyle \sum_{i \in J} c_i(v_i + N)=0$ implying $\displaystyle c_i = 0,$ for all $\displaystyle i,$ because the elements of $\displaystyle A$ are linearly independent. thus the elements of $\displaystyle \{v_i + M: \ i \in I \}$ are linearly independent and so $\displaystyle W = span \{v_i + M : \ i \in I\}$ is a subspace of $\displaystyle V/M$ isomorphic to $\displaystyle V/N$.

6. ## Re: Third Iso Theorem

well, the way i was looking at it was: suppose we define φ:V/N → V/M by φ(v+N) = v+M. φ is well-defined because M ⊆ N.

of course, in this case φ(v+N) = 0 (that is, 0+M) implies v is in M, which in turn implies v is in N, so ker(φ) = N, so we have an isomorphism between φ(V/N) and V/N.

the similarity being, the construction of φ (do we really need a basis? oh, and alex, if you're reading this, in light of what i wrote in the other thread, the fact that i'm asking this question here surely must be somewhat...ironic...)

7. ## Re: Third Iso Theorem

Originally Posted by Deveno
well, the way i was looking at it was: suppose we define φ:V/N → V/M by φ(v+N) = v+M. φ is well-defined because M ⊆ N.

of course, in this case φ(v+N) = 0 (that is, 0+M) implies v is in M, which in turn implies v is in N, so ker(φ) = N, so we have an isomorphism between φ(V/N) and V/N.

the similarity being, the construction of φ (do we really need a basis? oh, and alex, if you're reading this, in light of what i wrote in the other thread, the fact that i'm asking this question here surely must be somewhat...ironic...)
your map,$\displaystyle \phi,$ is not well-defined.

8. ## Re: Third Iso Theorem

Originally Posted by NonCommAlg
your map,$\displaystyle \phi,$ is not well-defined.
go on...

9. ## Re: Third Iso Theorem

if $\displaystyle v \in N \setminus M,$ then $\displaystyle v + N = 0$ but $\displaystyle \phi(v+N) = v + M \neq 0.$ for $\displaystyle \phi$ to be well-defined we need to have $\displaystyle N \subseteq M$ not $\displaystyle M \subseteq N.$

10. ## Re: Third Iso Theorem

hmm...so you're saying choosing the $\displaystyle v_i + N$ solves this problem, so we DO need a basis. huh. LIB!MR dux.

11. ## Re: Third Iso Theorem

Originally Posted by NonCommAlg
define the map $\displaystyle \phi: G/H \longrightarrow G/K$ by $\displaystyle \phi(gH)=gK,$ which is well-defined because $\displaystyle H \subseteq K$. clearly $\displaystyle \phi$ is a group homomorphism and $\displaystyle \ker \phi = K/H$. thus, by the first isomorphism theorem, $\displaystyle K/H$ is a normal subgroup of $\displaystyle G/H$ and $\displaystyle (G/H)/(K/H) \cong G/K$.
Is the whole proof really this compact?

12. ## Re: Third Iso Theorem

yup.

basically, the general strategy for proving G/K is isomorphic to G' is to find a surjective homomorphism G-->G', whose kernel is K, and then apply the FIT.

13. ## Re: Third Iso Theorem

Originally Posted by Deveno
yup.

basically, the general strategy for proving G/K is isomorphic to G' is to find a surjective homomorphism G-->G', whose kernel is K, and then apply the FIT.
Why is the kernel K\H?

14. ## Re: Third Iso Theorem

suppose φ(gH) = K, the identity of G/K. since φ(gH) = gK, this means g is in K. so gH is an element of K/H, which are the cosets of H in K.

so ker(φ) ⊆ K/H.

on the other hand, suppose we have kH as an element of G/H (so kH is a coset of H in G, where k is an element of K).

then φ(kH) = kK = K, so every element of G/H of the form kH for k in K, is in ker(φ). thus K/H ⊆ ker(φ), so the two sets are equal.

(remember, K/H is a subset (even stronger, a subgroup) of G/H consisting of only those elements kH, where k is in K).