On the maximal subgroup
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Let $G$ be a group, $M$ be a maximal subgroup of $G$ and $alpha in operatornameAut(G)$.
I want to show that $alpha(M)$ is a maximal subgroup of $G$.
I know $alpha(M)= lbrace alpha(m) mid m in M rbrace$. Suppose contrary that $alpha(M) <K<G$. please help me to complete this proof.
abstract-algebra
edited Aug 20 at 7:32
Bernard
111k635103
asked Aug 20 at 6:39
Hana
837
add a comment |Â
up vote
4
down vote
favorite
Let $G$ be a group, $M$ be a maximal subgroup of $G$ and $alpha in operatornameAut(G)$.
I want to show that $alpha(M)$ is a maximal subgroup of $G$.
I know $alpha(M)= lbrace alpha(m) mid m in M rbrace$. Suppose contrary that $alpha(M) <K<G$. please help me to complete this proof.
abstract-algebra
edited Aug 20 at 7:32
Bernard
111k635103
asked Aug 20 at 6:39
Hana
837
2
Hint: $M<alpha^-1(K)<G$, but $M$ is supposed to be maximal.
– Ãngel Mario Gallegos
Aug 20 at 6:52
@ Ãngel Mario Gallegos Thanks i get it ;)
– Hana
Aug 20 at 6:56
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $G$ be a group, $M$ be a maximal subgroup of $G$ and $alpha in operatornameAut(G)$.
I want to show that $alpha(M)$ is a maximal subgroup of $G$.
I know $alpha(M)= lbrace alpha(m) mid m in M rbrace$. Suppose contrary that $alpha(M) <K<G$. please help me to complete this proof.
abstract-algebra
edited Aug 20 at 7:32
Bernard
111k635103
asked Aug 20 at 6:39
Hana
837
Let $G$ be a group, $M$ be a maximal subgroup of $G$ and $alpha in operatornameAut(G)$.
I want to show that $alpha(M)$ is a maximal subgroup of $G$.
I know $alpha(M)= lbrace alpha(m) mid m in M rbrace$. Suppose contrary that $alpha(M) <K<G$. please help me to complete this proof.
abstract-algebra
edited Aug 20 at 7:32
Bernard
111k635103
asked Aug 20 at 6:39
Hana
837
edited Aug 20 at 7:32
Bernard
111k635103
edited Aug 20 at 7:32
Bernard
111k635103
edited Aug 20 at 7:32
Bernard
111k635103
111k635103
asked Aug 20 at 6:39
Hana
837
asked Aug 20 at 6:39
Hana
837
asked Aug 20 at 6:39
Hana
837
837
2
Hint: $M<alpha^-1(K)<G$, but $M$ is supposed to be maximal.
– Ãngel Mario Gallegos
Aug 20 at 6:52
@ Ãngel Mario Gallegos Thanks i get it ;)
– Hana
Aug 20 at 6:56
add a comment |Â
2
Hint: $M<alpha^-1(K)<G$, but $M$ is supposed to be maximal.
– Ãngel Mario Gallegos
Aug 20 at 6:52
@ Ãngel Mario Gallegos Thanks i get it ;)
– Hana
Aug 20 at 6:56
2
2
Hint: $M<alpha^-1(K)<G$, but $M$ is supposed to be maximal.
– Ãngel Mario Gallegos
Aug 20 at 6:52
Hint: $M<alpha^-1(K)<G$, but $M$ is supposed to be maximal.
– Ãngel Mario Gallegos
Aug 20 at 6:52
@ Ãngel Mario Gallegos Thanks i get it ;)
– Hana
Aug 20 at 6:56
@ Ãngel Mario Gallegos Thanks i get it ;)
– Hana
Aug 20 at 6:56
add a comment |Â
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2
Hint: $M<alpha^-1(K)<G$, but $M$ is supposed to be maximal.
– Ãngel Mario Gallegos
Aug 20 at 6:52
@ Ãngel Mario Gallegos Thanks i get it ;)
– Hana
Aug 20 at 6:56