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I am solving this problem:

Let $H,~K$, and $L$ be normal subgroups of $G$ with $H<K<L$. Let $A=G/H,~B=K/H$, and $C=L/H$

Show that $B$ and $C$ are normal subgroups of $A$, and $B<C$

My solution is:

Let us use the third isomorphism theorem: $$(G/H)/(K/H)\simeq G/K$$ Replacing: $$ A/B\simeq G/K$$ As the isomorphism preserves the structure of the group, then $B$ is normal to $A$. Also, $A/B$ would not be a group if $B$ was not normal in $A$.

An Analogous argument shows that $C$ is normal to $A$.

It follows from $H<K<L$ that $K/H<L/H$ so $B<C$.

Is my first argument correct? I have found another solution that uses the restricted natural homomorphisms, but I think it is unnecessarily confusing.

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This is sometimes considered part of the Correspondence Theorem.

I don't think your argument holds, because you would normally prove first that $B$ is normal in $A$ then $A/B\cong G/K$, precisely because "$A/B\cong G/K$" only makes sense if $B$ is normal in $A$. I'd say that what you are trying to prove is usually proved before the third isomorphism theorem, so you should prove it without using the third isomorphism theorem.

You should think to yourself 'what does it mean for $B$ to be normal in $A$?' then prove it. So, $B$ is normal in $A$ if and only if $\forall x\in K$, $g\in G$ we have $(gH)^{-1}xHgH=xH$. You should be able to write prove this does hold in one or two steps. The analogous argument holds for $C$.

As for $B<C$, you seem to be restating what you are trying to prove rather than proving it. Honestly the statement is so obvious, I think most people would do the same and in an exam / assignment the marker might be generous, but as you are asked to prove it you should spell it out: 'If $X\in B$ then $X=xH$ for some $x\in K\le L$ so $x\in L$ and $X\in C$'.

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Actually the quotient K/H is also a normal subgroup is in the statement of third isomorphism theorem[1].So, It would be circular argument to use theorem its self to prove the theorem

[1]Dummit, David S.; Foote, Richard M., Abstract algebra, Wiley International Edition. Chichester: Wiley (ISBN 0-471-45234-3/hbk). xii, 932 p. (2004). ZBL1037.00003 page 98.

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