If $G_1\cong G_2$, $H_1\triangleleft G_1$, $H_2 \triangleleft G_2$ and $G_1/H_1\cong G_2/H_2$, then is $H_1\cong H_2$?

I am a new student in a learning group.

Assume $$G_1$$, $$G_2$$ are two groups, $$H_1\triangleleft G_1$$, $$H_2\triangleleft G_2$$.

1. 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： $$G_1=G_2=\mathbb{Z}，~H_1=2\mathbb{Z},~H_2=3\mathbb{Z}.$$

1. 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!

• Third condition is false. Commented Aug 12, 2022 at 11:40

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).