Let $A,B\triangleleft G$. Give counterexample for the claims:
a. $G/A\cong B \Rightarrow G/B\cong A$
b. $G/A\cong G/B\Leftrightarrow A\cong B$
I don't know from where to start. Can you give some counterexamples? addiotionally, what is the intuition of solving question in algebra which requires giving counter-examples
Hint: Take $G=\mathbb{Z}_4\times \mathbb{Z}_2$. This group provides a basis for counter-examples to both of your statements.
Assuming that $G/_A$ is the same as $G/A$; for $\bf b$ we can see that:
$A=2\mathbb Z\cong 3\mathbb Z=B$ but $G/A=\mathbb Z_2$ is not isomorphic with $G/B=\mathbb Z_3$ of order $3$.
If $G=D_8=\langle a,b\mid a^4=b^2=e, ba=a^3b\rangle$, the dihedral group of order $8$, and $A=\langle a\rangle,~~~ B=\{e,a^2,b,a^2b\}$ then $G/A\cong G/B$.
There is no "intuition of solving question in algebra which requires giving counter-examples"; what there is, is knowing lots of examples so you can test them for counterexamples. You should know, for example, all the finite abelian groups, the symmetric, alternating, and dihedral groups, and the quaternion group, and also how to construct groups by taking direct products.
Now, for a), assuming the groups are finite, you need an order where there are two nonisomorphic groups, and you should have sufficient familiarity with small groups to know that the smallest example is that there are two groups of order 4. Let's call them $X$ and $Y$. So you can look for a group $G$ with subgroups $X$, $Y$, and $Z$, such that $G/X$ is isomorphic to $Z$, while $G/Z$ is isomorphic to $Y$. And try to take $Z$ as small as possible, which would be order 2. That would make $G$ order 8. That should tell you where to look.
