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Show that $(\Bbb Z/15\Bbb Z)^{\times}\simeq\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z$, where $(\Bbb Z/15\Bbb Z)^{\times}$ is the group of integers modulo $15$ under multiplication.

This is a question involving the First Isomorphism Theorem but I don't know how to use it with a direct product. I've checked whether the groups are cyclic and also tried to just find functions $f:\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\to(\Bbb Z/15\Bbb Z)^{\times}$ but that didn't get me anywhere. If possible, a hint would help.

cqfd
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CS24B
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    How many elements are there in $(\Bbb Z/15\Bbb Z)^\times$? Use this together with the fundamental theorem of finite abelian groups. – LetGBeTheGraph Aug 18 '20 at 13:43

1 Answers1

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We always have $$ (\Bbb Z/pq\Bbb Z)^{\times}\cong (\Bbb Z/p\Bbb Z)^{\times}\times (\Bbb Z/q\Bbb Z)^{\times}, $$ for primes $p$ and $q$ by the CRT (Chinese Remainder Theorem).

Furthermore we have $(\Bbb Z/p\Bbb Z)^{\times}\cong \Bbb Z/(p-1)\Bbb Z$.

References:

$\mathbb Z_{mn}$ isomorphic to $\mathbb Z_m\times\mathbb Z_n$ whenever $m$ and $n$ are coprime

Is my proof that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes correct?

Dietrich Burde
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