Let $G=\langle a\rangle$ and $G'=\langle b\rangle$ of order 12 and 4. Prove that there is a onto homomorphism with $o(Ker~ \phi)=3$.

Question

Let $G=\langle a\rangle$ and $G'=\langle b\rangle$ be two cyclic groups of order 12 and 4 respectively. Prove that there exists a onto homomorphism $\phi: G \rightarrow G'$ with $o(Ker~ \phi)=3$.

Attempt:

$G=\{e_G, a, a^2, \cdots a^{11}\}$ and $o(a)=12=o(G)$

$G'=\{e_{G'}, b, b^2, b^{3}\}$ and $o(b)=4=o(G')$

Clearly, as $4|12$, there exists a subgroup $H$ of $G$ of order 3 generated by $a^m$ where $m=12/3=4$.

Isomorphism theorem gives $G/{Ker~ \phi}\simeq G'$ if $\phi$ is an onto homomorphism. Then

1. How to prove that $\phi$ is an onto homomorphism?

2. How to show $o(Ker~ \phi)=3$

Please help me in an elaborate form so that I can understand. Thanks.

Edit 1:

If $\phi$ is an onto homomorphism then I can find $o(Ker~ \phi)$ as follows

I can apply Isomorphism Theorem $G/{Ker~ \phi}\simeq G'$ then $o( G/{Ker~ \phi})=o( G')$ i.e $$\frac{o( G)}{o({Ker~ \phi})}=o(G')$$ that is $$\frac{12}{o({Ker~ \phi})}=4$$ that is $$o({Ker~ \phi})=12/4=3.$$

How to solve 1st part. Please help.

Edit 2:

How to show that it is Onto Homomorphism for this particular problem?

If $e$ and $e′$ are the identity elements of $G$ and $G′$ respectively.

For any $x\in G$ then $ϕ(x)=ϕ(ex)=ϕ(e+e+⋯+e)=ϕ(e)+ϕ(e)+⋯+ϕ(e)=xϕ(e)$. Then $ϕ$ is completly known if $ϕ(e)$ is known.

Answer 1

Define $\phi(a^n) = b^n$ for all $n \in \Bbb Z$. Let us show that this is a morphism.

If $g = a^m \in G$ and $h = a^n \in G$, then

$$\phi (gh) = \phi (a^m a^n) = \phi (a^{m+n}) = b^{m+n} = b^m b^n = \phi (a^m) \phi (a^n) = \phi (g) \phi (h) .$$

If $g = a^m$, then

$$\phi (g^{-1}) = \phi ((a^m)^{-1}) = \phi (a^{-m}) = b^{-m} = (b^m)^{-1} = \phi (g) ^{-1} .$$

Finally, letting $n=0$ in the definition, we get

$$\phi (e) = \phi (a^0) = b^0 = e' .$$

Thus, $\phi$ is a group morphism.

Notice now that if $a^m \in \ker \phi$, then $b^m = e'$ and since $b$ is a generator of $G'$, it follows that $\text{ord } b \mid m$, which means $4 \mid m$. Conversely, if $m = 4k$, then $\phi (a^m) = \phi (a^{4k}) = b^{4k} = (b^4)^k = e' ^k = e'$. This shows that $\ker \phi = \{ a^{4k} \mid k \in \Bbb Z \} = \{e, a^4, a^8\}$ (because $a^{12} = e$), which is a subgroup with three elements, hence of order $3$.

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To show that $\phi$ is onto, use the fundamental isomorphism theorem and compare the cardinals:

$$\left| \frac {G} {\ker \phi} \right| = |\phi (G')| ,$$

and by Lagrange's theorem the left-hand side is just $\dfrac {12} 3 = 4$, which is just $|G'|$, whence it follows that $\phi (G) = G'$, so $\phi$ is onto.

Alternatively (and more explicitly), if $b^n \in G'$ and $n = 4q + r$ with $r \in \{0,1,2,3\}$, then

$$\phi ( a ^r) = b^r = e' b^r = (b^4)^q b^r = b^{4q + r} = b^n ,$$

so $\phi$ is onto.

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Honestly, the whole proof is a bit long when done with all the details as above. The idea, though, is easy: define $\phi(a) = b$ (i.e. send the generator of $G$ to the generator of $G'$) and extend this by multiplication.

Answer 2

Hint: what should $\phi(a)$ be?