# Onto Homomorphism to cyclic group

## Homework Statement

If P: G-->C_6 is an onto group homomorphism and |ker(p)| = 3, show that |G| = 18 and G has normal subgroups of orders 3, 6, and 9.

C_6 is a cyclic group of order 6.

none

## The Attempt at a Solution

I determined that |G| = 18 by taking the factor group G/Ker(p) and realizing it is isormorphic to C_6 and so they must have equal order, then using Lagranges law (I think) I solved for |G| = |Ker(p)|*|C_6| = 18.

G must have a normal subgroup of order 3 because Ker(p) is a normal subgroup of G with order 3. How do I show that G has normal subgroups of order 6 and 9? Do I find a way to show that homomorphisms exist where the order of the kernals are 6 and 9?

Dick
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## Homework Statement

If P: G-->C_6 is an onto group homomorphism and |ker(p)| = 3, show that |G| = 18 and G has normal subgroups of orders 3, 6, and 9.

C_6 is a cyclic group of order 6.

none

## The Attempt at a Solution

I determined that |G| = 18 by taking the factor group G/Ker(p) and realizing it is isormorphic to C_6 and so they must have equal order, then using Lagranges law (I think) I solved for |G| = |Ker(p)|*|C_6| = 18.

G must have a normal subgroup of order 3 because Ker(p) is a normal subgroup of G with order 3. How do I show that G has normal subgroups of order 6 and 9? Do I find a way to show that homomorphisms exist where the order of the kernals are 6 and 9?

Write down a homomorphism Q:C_6-->C_6 such that ker(Q) has order 2. Suppose you consider the composition QP?

Last edited:
• PsychonautQQ
jbunniii
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G must have a normal subgroup of order 3 because Ker(p) is a normal subgroup of G with order 3. How do I show that G has normal subgroups of order 6 and 9? Do I find a way to show that homomorphisms exist where the order of the kernals are 6 and 9?
Hint: ##C_6## has normal subgroups of orders 1, 2, and 3. (Why?) Now apply the correspondence theorem.

• PsychonautQQ