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Suppose that $G$ is a group of order 5 and let $\varphi: \mathbb{Z_{30}} \rightarrow G$ be and epimorphism. Determine the kernel of $\varphi$

Since $\ker\varphi \unlhd \mathbb{Z_{30}}$ (theorem) then since $\ker\varphi$ is a subgroup of $\mathbb{Z_{30}}$ by lagranges theorem $[\mathbb{Z_{30}}:\ker\varphi]$ = $|\mathbb{Z_{30}}|$/$|\ker\varphi|$

Since $|G| = 5$ and $|\mathbb{Z_{30}}| = 30$, would the $[\mathbb{Z_{30}}:\ker\varphi] = 30/5=6$ and thus $|\ker\varphi| =5$ so that the $\ker\varphi = \{5,10,15,20,25\}$?

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  • $\begingroup$ It would mean $30/|\ker\varphi|=5$ not what you wrote. Also $0\in\ker$ always. $\endgroup$
    – anon
    Commented Apr 20, 2014 at 18:24

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This is almost correct: $0$ is always an element of the kernel of a homomorphism.

Note that by the first isomorphism theorem, we actually have that

$$|G| = \frac{|\mathbb{Z}_{30}|}{|\ker \varphi|} \implies |\ker \varphi| = \frac{30} 5 = 6$$

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  • $\begingroup$ Oh you're right, I didn't even see that from the first Iso theorem. Makes so much more sense now!! $\endgroup$
    – spitfiredd
    Commented Apr 20, 2014 at 18:27
  • $\begingroup$ $|G|=5$ by hypothesis so the OP needs to find their error. $\endgroup$
    – anon
    Commented Apr 20, 2014 at 18:32

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