# group homomorphism and subgroups

#### alberto

##### New member
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
$$\phi : G \to G'$$
is a homomorphism
which means for every $$g_1,g_2 \in G$$
$$\phi(g_1*_{G} g_2 ) = \phi (g_1 ) *_{G'} \phi(g_2)$$

H' is a subgroup of G'
$$H=\phi ^{-1} ( H' )$$

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

#### caffeinemachine

##### Well-known member
MHB Math Scholar
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.