An exercise with the third isomorphism theorem in group theory

[Alex Langevub]

Homework Statement

Let $G$ be a group. Let $H \triangleleft G$ and $K \leq G$ such that $H\subseteq K$.

a) Show that $K\triangleleft G$ iff $K/H \triangleleft G/H$

b) Suppose that $K/H \triangleleft G/H$. Show that $(G/H)/(K/H) \simeq G/K$

Homework Equations

The three isomorphism theorems.

The Attempt at a Solution

a) Let $k \in K$, $h \in H$ and $g \in G$

$(gH)(kH)(gH)^{-1}$ need to belong to $kH$
I am not quite sure how to go about proving this.

b) Since $K/H \triangleleft G/H$, we know that $K\triangleleft G$ as proven in a).

We can therefore apply the third isomorphism theorem which states that since $H$ and $K$ are both normal in $G$ and that $H$ is a subset of $K$,
$$(G/H)/(K/H) \simeq (G/K)$$

 

[fresh_42]

Homework Statement

Let $G$ be a group. Let $H \triangleleft G$ and $K \leq G$ such that $H\subseteq K$.

a) Show that $K\triangleleft G$ iff $K/H \triangleleft G/H$

b) Suppose that $K/H \triangleleft G/H$. Show that $(G/H)/(K/H) \simeq G/K$

Homework Equations

The three isomorphism theorems.

The Attempt at a Solution

a) Let $k \in K$, $h \in H$ and $g \in G$

$(gH)(kH)(gH)^{-1}$ need to belong to $kH$
I am not quite sure how to go about proving this.

You will have to use that $K$ is normal in $G$, so $h^{-1}kh=k'$ and you can always write $H=Hh^{-1}$. Proceed elementwise, just that you cannot assume the same $h$ as representative of $H$. Every occurrence has another $h_i\,.$ And don't forget the other direction!

b) Since $K/H \triangleleft G/H$, we know that $K\triangleleft G$ as proven in a).

We can therefore apply the third isomorphism theorem which states that since $H$ and $K$ are both normal in $G$ and that $H$ is a subset of $K$,
$$(G/H)/(K/H) \simeq (G/K)$$

I thought part b) is exactly the isomorphism theorem, which has to be shown, but if you have it already, then yes, your reasoning is correct.

 

[Alex Langevub]

You will have to use that $K$ is normal in $G$, so $h^{-1}kh=k'$ and you can always write $H=Hh^{-1}$. Proceed elementwise, just that you cannot assume the same $h$ as representative of $H$. Every occurrence has another $h_i\,.$ And don't forget the other direction!

I thought part b) is exactly the isomorphism theorem, which has to be shown, but if you have it already, then yes, your reasoning is correct.

I am still unsure how to go about resolving this problem. I haven't really seen any examples of problems with quotients like this one. So it's the possible manipulations that I am unsure about. Please tell me if this is alright.This problem is an iif so I need to demonstrate both directions ⇐) and ⇒).⇒)
If $K\triangleleft G$, so we have that $g_1^{-1}k_1g_1 \in K$.

$$(g_1h_1)(k_2h_2)(g_1h_1)^{-1}$$
lets set $k_2 = g_1^{-1}k_1g_1$
$$(g_1h_1g_1^{-1})k_1(g_1h_2h_1^{-1}g_1^{-1}) $$

I am not sure where to go from there. Or even if I am on the right track...

 

[fresh_42]

Let me see.

$(gh_1)(kh_2)(g^{-1}h_3)=gh_1kh_1^{-1}h_1h_2g^{-1}h_3=gk\,'h_1h_2g^{-1}h_3=gk\,'g^{-1}gh_1h_2g^{-1}h_3=k\,''gh_1h_2g^{-1}h_3 =kh_4h_3 \in KH$ because both $H$ and $K$ are normal.

Now the other direction: $K/H \trianglelefteq G/H \Longrightarrow K \trianglelefteq G\,.$

 

[Alex Langevub]

I assume you meant $kh_4h_3 \in K/H$ and not $kh_4h_3 \in KH$ at the end there.Now,
$\Leftarrow$

In an exam I would put a lot more effort into initializeg the different variables $k_i$, $g_i$ and $h_i$ and justifying each step.

If $K/H \triangleleft G/H$, we have that
$(gh_1)(k_1h_2)(g^{-1}h_3) = k'h'$
$gh_1k_1g^{-1}(gh_2g^{-1})h_3 = k'h'$
$g(h_1k_1)g^{-1}(h_4)h_3 = k'h'$
$g(k_2)g^{-1}h_5 = k'h'$
$\Rightarrow gk_2g^{-1} \in K$
$\Rightarrow K\triangleleft G$

 

[fresh_42]

I assume you meant $kh_4h_3 \in K/H$ and not $kh_4h_3 \in KH$ at the end there.

No. I meant what I wrote, except that I forgot to write $k''$ instead of $k$. We still have elements: $(gh_1)(kh_2)(g^{-1}h_3)=k''h_4h_3 \in KH$ but from that passing to cosets yields $[g][k][g]^{-1}=[k''] \in K/H$ what we needed.

Now,
$\Leftarrow$

In an exam I would put a lot more effort into initializeg the different variables $k_i$, $g_i$ and $h_i$ and justifying each step.

If $K/H \triangleleft G/H$, we have that

Let me fit in some lines which helps me to sort stuff out. We have $H \triangleleft G$ and thus also $H \triangleleft K$.
Now we have $(gH)(k_1H)(g^{-1}H) \in K/H\,,$ so

$(gh_1)(k_1h_2)(g^{-1}h_3) = k'h'$
$gh_1k_1g^{-1}(gh_2g^{-1})h_3 = k'h'$
$g(h_1k_1)g^{-1}(h_4)h_3 = k'h'$
$g(k_2)g^{-1}h_5 = k'h'$
$\Rightarrow gk_2g^{-1} \in K$

$\Rightarrow gk_2g^{-1} = k'h_6 \in KH \subseteq K$

$\Rightarrow K\triangleleft G$

O.k.