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Let $G$ be a finite group and assume $\varphi: \, G \longrightarrow \mathbb{Z}_{10}$ is a group epimorphism. I want to show that there exist an $a \in G$ s.t. $\text{ord}(a)=5$.

At first, the cyclic group: $$ \langle2\rangle=\{0,2,4,6,8\} $$ is a subgroup of $\mathbb{Z_{10}}$ of order $5$.

The epimorphism $\varphi$ induces anοther epimorphism $\varphi ': \langle a\rangle \longrightarrow \langle 2 \rangle$, where $a \in G$ s.t. $\varphi(a)=2$. By the fundamental homomorphism theorem: $$ \langle 2 \rangle \cong \langle a\rangle/\text{ker}\varphi '\iff |\langle a\rangle|=5 \cdot |\text{ker}\varphi '| $$ Is there a way to prove that $|\text{ker}\varphi '|=1$ for some $a \in G$? In other words, how can it be shown that there exists an $a \in G$ which makes $\varphi '$ an isomorphism?

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3 Answers 3

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There's a far easier option: since $\varphi$ is epic, it's surjective, so there is some $g \in G$ such that $\varphi(g) = 2$, and the order of $2$ in $\mathbb{Z}_{10}$ is $5$, so the order of $g$ in $G$ is a multiple of $5$ (there is some $n$ such that $g^n = 1_G$ since $G$ is finite, and if $n = 5a+b$, then $0 = \varphi(1_G) = \varphi(g^{5a+b}) = \varphi(g)^{5a+b} = 2^{5a+b} = 2^b$, so $b$, hence $n$, is a multiple of $5$).

So $|g| = 5k$ for some $k$, and $|g^k| = 5$.

You can do a similar thing with your approach: once you know that $\varphi'$ is an epimorphism, you know that $5$ divides the order of $a$, so essentially the same argument tells you that the order of some power of $a$ is $5$ (and just to answer your final question: yes, just raise your original $a$ to the power of $|\ker\varphi'|$).

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Note that $\mathbb{Z}/10 \cong G/K$ for some $K$ a normal subgroup of $G$. In particular, $|G| = 10|K|$, so $|G|$ is a multiple of $5$. Therefore by Cauchy's theorem, there exists an element of $G$ of order $5$.

I'm not sure how to proceed by your method (except by invoking Cauchy's theorem anyway) but will have a think.

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I think you are overcomplicating this. By First Isomorphism Theorem, $G/N\equiv\Bbb{Z}_{10}$, where $N$ is the kernel of $\phi$. However, this means the order of $G/N$ is $10$. Since the order of $G/N$ is a factor of the order of $G$, this means the order of $G$ is a multiple of $10$. Therefore, the order of $G$ is a multiple of $5$, since $5$ is a multiple of $10$. Thus, by Cauchy's theorem, $G$ contains an element of order $5$ because $5$ is prime.

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