Let \(H, K \leq G\) be subgroups with \(H\cong K\). Is it true that \(G/H \cong G/K\)?
Hint: consider a group with distinct subgroups of order 2 whose quotients have order 4.
Solution
No.
1. Let\[ G=C_4\oplus C_2, \]and write elements additively. Define\[ H=\langle(2,0)\rangle, \qquad K=\langle(0,1)\rangle. \]Then \(H\cong K\cong C_2\).
Both displayed generators have order \(2\). Since \(G\) is abelian, both subgroups are normal, so both quotients are defined.
2. The quotient \(G/H\) is isomorphic to \(C_2\oplus C_2\).
Quotienting the \(C_4\) factor by its subgroup of order \(2\) gives \(C_2\), while the second factor survives unchanged: \[ G/H\cong (C_4/\langle2\rangle)\oplus C_2\cong C_2\oplus C_2. \]
3. The quotient \(G/K\) is isomorphic to \(C_4\).
The subgroup \(K\) is exactly the second direct factor, so \[ G/K\cong C_4. \]
4. Hence \(G/H\not\cong G/K\).
The group \(C_4\) contains an element of order \(4\), whereas every nonzero element of \(C_2\oplus C_2\) has order \(2\). Thus the quotients are not isomorphic even though \(H\cong K\).