Suppose there is homomorphism from finite group G onto $Z_{10}$ Then G has Normal Subgroup of index 5 and 2.

Question

I had come across following Problem given by :
[I know here is same question But i am not convince with argument given By One of my Respected Teacher on MSE Prove $G$ has normal subgroups of indexes 2 and 5

Suppose there is homomorphism from finite group G onto $Z_{10}$ Then G has Normal Subgroup of index 5 and 2.

I know By First Fundamental Theorem of Isomorphism $G/Ker \phi$ $\approx$ $Z_{10}$
So $10\space|\space|G|$.As $Z_{10}$ is abelian every subgroup is normal so $\left<2\right>$ and $\left<5\right>$ are normal subgroup . $\phi^{-1}(\left<2\right>)$ and $\phi^{-1}(\left<5\right>)$ are normal in G. Order of $\phi^{-1}(\left<5\right>)$ is multiple of 2 i.e $2k$ and $\phi^{-1}(\left<2\right>)$ is $5m$
I had doubt that how this I can convert to show that they have required index normal group? Any Help Will be appreciated

Answer 1

Use the isomorphism theorem you have that $G/\ker \cong Z_{10}$. Let $\varphi:G/\ker \rightarrow Z_{10}$ be the isomorphism.

Let $H_5 = \left<5\right>$ and $H_2=\left<2\right>$, and let $G_1 := \varphi^{-1}(H_5)$ and $G_2:=\varphi^{-1}(H_2)$. We can lift $G_1,G_2$ into subgroups $\overline{G}_1,\overline{G}_2$ of $G$.

Lemma: Let $A,B$ be isomorphic groups, let $\tau:A\rightarrow B$ be the isomorphism and let $C\leq B$ be a subgroup. Then $\tau$ induces an isomorphism $A/\tau^{-1}(C)\rightarrow B/C$.

Applying this lemma in our case for $A=G/\ker$ and $B=Z_{10}$, we have that $G/\overline{G}_1 \cong Z_{10}/H_5 = Z_5$ and $G/\overline{G}_2 = Z_{10}/H_2$. Hence the indices of $\overline{G}_1,\overline{G}_2$ are $5,2$ respectively.

Edit: I add a proof for the lemma. We have

$\tau:A\rightarrow B$ an isomorphism. Let $\pi:B\rightarrow B/C$ be the projection map $b\mapsto b+C$. Consider the map $\pi\circ\tau:A\rightarrow B/C$ this is a surjective homomorphism. Apply the first isomorphism theorem we have that $A/\ker (\pi\circ\tau) \cong B/C$. It is thus left to show that $\ker(\pi\circ\tau) = \tau^{-1}(C)$.

Clearly, if $a\in \tau^{-1}(C)$ so $\tau(a)\in C$ and so $\pi(\tau(a))=1+C$ hence $\tau^{-1}(C)\subseteq \ker (\pi\circ\tau)$. On the other hand, if $\pi\circ\tau (a) = 1+C$ then $\tau(a)\in C$ (because $\pi\circ\tau(a)=\tau(a)+C$) it follows that $a\in \tau^{-1}(C)$, hence $\ker(\pi\circ\tau)\subseteq \tau^{-1}(C)$. This completes the proof.

Answer 2

Let $\varphi\colon G\to\mathbb{Z}_{10}$ be a surjective homomorphism. Take a subgroup $K$ of $\mathbb{Z}_{10}$ of index $k$ (with $k=2$ or $k=5$).

Then you can consider the projection map $\pi\colon\mathbb{Z}_{10}\to\mathbb{Z}_{10}/K$ and the composition $\psi=\pi\circ\varphi$. Then $\psi$ is a surjective homomorphism and so $$ G/\ker\psi\cong\mathbb{Z}_{10}/K $$ by the homomorphism theorem. Therefore $\ker\psi$ has index $k$, because $|\mathbb{Z}_{10}/K|=k$ by assumption.