I am a new student in a learning group.
Assume $G_1$, $G_2$ are two groups, $H_1\triangleleft G_1$, $H_2\triangleleft G_2$.
- If $G_1\cong G_2$, $H_1\cong H_2$, I know there is no $G_1/H_1\cong G_2/H_2$.
Counterexample： $$G_1=G_2=\mathbb{Z}，~H_1=2\mathbb{Z},~H_2=3\mathbb{Z}.$$
- If $G_1/H_1\cong G_2/H_2$, $H_1\cong H_2$, I know there is no $G_1\cong G_2$.
Counterexample： $$G_1=\mathbb{Z}_4=\langle g\rangle，G_2=\left\{1,a,b,ab|ab=ba,a^2=b^2=1\right\}, ~H_1=\langle g^2\rangle,~H_2=\langle a\rangle.$$
I want to know if the third condition is correct?
If $G_1\cong G_2$, $H_1\lhd G_1$ ,$H_2 \lhd G_2$ and $G_1/H_1\cong G_2/H_2$, then is $H_1\cong H_2$?
Thanks!
Best Answer
The third condition is NOT correct.
Consider $D_8$, the dihedral group of order 8, which represents the symmetries of a square in the plane. A presentation for $D_8$ is $$D_8 = \langle r, s ~|~ r^4=1, s^2 =1, sr = r^3s \rangle$$ where $r$ represents a rotation by 90 degrees and $s$ represents a reflection.
Take $G_1 = G_2 = D_8$. Take $H_1 = \langle r\rangle$ which consists of $\{1, r, r^2, r^3\}$ and take $H_2 = \langle r^2, s\rangle$ which consists of $\{1, r^2, s, r^2s\}$. Note that $H_1$ and $H_2$ are both subgroups of index 2 in $D_8$, hence are both normal subgroups and their quotient groups are both isomorphic to $\mathbb Z/2\mathbb Z$. But $H_1$ is isomorphic to $\mathbb Z/4\mathbb Z$ (since it is generated by an element of order 4) and and $H_2$ is isomorphic to $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$ (since all its elements are of order 2).