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group homomorphism and subgroups

alberto

New member
Mar 24, 2012
1
Hi,
I am having trouble with this question so it would be really nice if anyone could provide some help.

Let $$\phi: G \to G'$$ be a group homomorphism, and let $$H' \le G'$$ be a subgroup of G'.

a) Show that $$H=\phi^{-1}(H')$$ is a subgroup of G.
b) Now suppose H’ is a normal subgroup of G’. Does it follow that $$H=\phi^{-1}(H’)$$ is a normal subgroup of G?
c) Now suppose phi is surjective (but not that H’ is normal). Show that there is a bijective correspondence between $$K’ \le G’$$ containing H' and $$K \le G$$ containing H.

Thanks
 

Amer

Active member
Mar 1, 2012
275
[tex] \phi : G \to G' [/tex]
is a homomorphism
which means for every [tex]g_1,g_2 \in G [/tex]
[tex]\phi(g_1*_{G} g_2 ) = \phi (g_1 ) *_{G'} \phi(g_2) [/tex]

H' is a subgroup of G'
[tex]H=\phi ^{-1} ( H' ) [/tex]

the subgroup text is if [tex]h_1,h_2 \in H \; [/tex] then [tex]h_1h_2^{-1} \in H [/tex]
so let [tex]h_1 , h_2 \in H [/tex] then there exist [tex] h_1 ' , h_2 ' \in H ' [/tex]
such that [tex]\phi^{-1} (h_1 ' ) = h_1 , \phi^{-1} (h_2 ' ) = h_2 [/tex]
[tex]\phi(h_1 ) = h_1 ' , \phi(h_2 ) = h_2 ' [/tex]
[tex]\phi(h_1h_2^{-1}) = \phi(h_1)(\phi(h_2))^{-1} = h_1' (h_2')^{-1} \in H [/tex]
[tex]h_1h_2^{-1} \in \phi^{-1}(H) = H [/tex]
the proof ends
note that from the homomorphism
[tex]\phi(h^{-1}) = ( \phi( h ) )^{-1}[/tex]
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi,
I am having trouble with this question so it would be really nice if anyone could provide some help.

Let $$\phi: G \to G'$$ be a group homomorphism, and let $$H' \le G'$$ be a subgroup of G'.

a) Show that $$H=\phi^{-1}(H')$$ is a subgroup of G.
b) Now suppose H’ is a normal subgroup of G’. Does it follow that $$H=\phi^{-1}(H’)$$ is a normal subgroup of G?
c) Now suppose phi is surjective (but not that H’ is normal). Show that there is a bijective correspondence between $$K’ \le G’$$ containing H' and $$K \le G$$ containing H.

Thanks
Well.. I think you should submit your attempt along with the question too. That way you get help exactly where you are stuck and you get a lot more people willing to help you.
part a) Apply the "One step subgroup test". Let $a,b \in \phi^{-1}(H{'})$. Then $\phi (a), \phi (b) \in H{'}$. Thus $\phi (a), \phi (b^{-1}) \in H{'}$ (why?)
thus $\phi(ab^{-1}) \in H{'}$ and hence $ab^{-1} \in \phi ^{-1} (H{'})$. So by the one step subgroup test we have $\phi ^{-1} (H{'})$ is a subgroup of $G$.

For part b) and c) please show your attempt. If you are totally clueless on something that's fine too but do mention it.