- #1
mathmari
Gold Member
MHB
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Hey!
I want to show that if $H$ is a cylic normal subgroup of a group $G$, then each subgroup of $H$ is a normal subgroup of $G$.
I have done the following:
Since $H$ is a normal subgroup of $G$, we have that $$ghg^{-1}=h\in H, \ \forall g \in G \text{ and } \forall h\in H \tag 1$$
Since $H$ is cyclic, each subgroup of $H$ is also cyclic.
Let $\langle a\rangle$ be a subgroup of $H$.
Then $a\in \langle a\rangle \Rightarrow a\in H$.
From $(1)$ we have that $$gag^{-1}=a\in \langle a\rangle , \ \forall g\in G \text{ and } \forall a\in \langle a\rangle$$
Since the subgroup was arbitrary, we have that each subgroup of $H$ is a normal subgroup of $G$. Is everything correct? (Wondering)
I want to show that if $H$ is a cylic normal subgroup of a group $G$, then each subgroup of $H$ is a normal subgroup of $G$.
I have done the following:
Since $H$ is a normal subgroup of $G$, we have that $$ghg^{-1}=h\in H, \ \forall g \in G \text{ and } \forall h\in H \tag 1$$
Since $H$ is cyclic, each subgroup of $H$ is also cyclic.
Let $\langle a\rangle$ be a subgroup of $H$.
Then $a\in \langle a\rangle \Rightarrow a\in H$.
From $(1)$ we have that $$gag^{-1}=a\in \langle a\rangle , \ \forall g\in G \text{ and } \forall a\in \langle a\rangle$$
Since the subgroup was arbitrary, we have that each subgroup of $H$ is a normal subgroup of $G$. Is everything correct? (Wondering)