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mathmari
Gold Member
MHB
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Hey!
I am facing some difficulties understanding the solution of the exercise:
"Let $G_1,G_2$ groups. $|G_1|=15, |G_2|=56$
Show that it is not possible that there exist an isomorphism $f:G_1 \rightarrow G_2$."
which is the following:
($<$ means subgroup)
($e_1$:identity element of $G_1$, $e_2$:identity element of $G_2$)
$H_1=f^{-1}(G_2)<G_1$ from the property: If $H_2<G_2$, then $f^{-1}(H_2)<G_1$
$H_1$ is a subgroup of $G_1$ $\displaystyle{\overset{\text{Lagrange}}{\Rightarrow} |H_1| \mid 15}$
So $|H_1| \in \{1,3,5,15\}$
If $|H_1|=1$, that means that $H_1=\{e_1\}$, that means that $f^{-1}(G_2)=\{a \in G_1: f(a) \in G_2\}=\{e_1\}$.
But $f$ is onto, so there are $a \neq e_1$ with $f(a) \neq e_2$.
If it were $f^{-1}(G_2)=\{e_1\}$ that would mean that all $f(a)=e_1$.
So $|H_1| \in \{3,5,15\}$
Since $H_1<G_1$, $f(H_1)<G_2$, so $|f(H_1)| \mid |G_2|$
Since $f$ is $1-1$, $|f(H_1)|=3,5,15$
$|G_2|=56$ but none of the $3,5,15$ divide $56$.
Could you explain me the red part?? Why does this stand?? (Wondering)
I am facing some difficulties understanding the solution of the exercise:
"Let $G_1,G_2$ groups. $|G_1|=15, |G_2|=56$
Show that it is not possible that there exist an isomorphism $f:G_1 \rightarrow G_2$."
which is the following:
($<$ means subgroup)
($e_1$:identity element of $G_1$, $e_2$:identity element of $G_2$)
$H_1=f^{-1}(G_2)<G_1$ from the property: If $H_2<G_2$, then $f^{-1}(H_2)<G_1$
$H_1$ is a subgroup of $G_1$ $\displaystyle{\overset{\text{Lagrange}}{\Rightarrow} |H_1| \mid 15}$
So $|H_1| \in \{1,3,5,15\}$
If $|H_1|=1$, that means that $H_1=\{e_1\}$, that means that $f^{-1}(G_2)=\{a \in G_1: f(a) \in G_2\}=\{e_1\}$.
But $f$ is onto, so there are $a \neq e_1$ with $f(a) \neq e_2$.
If it were $f^{-1}(G_2)=\{e_1\}$ that would mean that all $f(a)=e_1$.
So $|H_1| \in \{3,5,15\}$
Since $H_1<G_1$, $f(H_1)<G_2$, so $|f(H_1)| \mid |G_2|$
Since $f$ is $1-1$, $|f(H_1)|=3,5,15$
$|G_2|=56$ but none of the $3,5,15$ divide $56$.
Could you explain me the red part?? Why does this stand?? (Wondering)
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