Let $(\Bbb{Z}_6,+)$ and $(\Bbb{Z}_5,\cdot)$ be two groups.

Define an operation for $(\Bbb{Z}_6,+)\times(\Bbb{Z}_5,\cdot)$ such that $(\Bbb{Z}_6,+)\times(\Bbb{Z}_5,\cdot)$ is a group. Find the identity and inverses of the product.

I know the following:

Property If $(G_1,*_1)$ and $(G_2,*_2)$ are two GROUPS, then the set $G_1\times G_2$ with the operation $(a,b)*(c,d)=(a*_1c,b*_2d)$ is also a GROUP.

But here, $(\Bbb{Z}_5,\cdot)$ is NOT a group since the inverse element of $0$ does not exist. So what should we do here?

**Should we consider another operation like $$(a,b)*(c,d)=(a+_6c,b+_5d)?$$

But the hypothesis is not true because $(\Bbb{Z}_5,\cdot)$ is not a group, so I do not know if I can define such operation of the product of "two" groups.

Any help?

  • $\begingroup$ Perhaps the author of the question meant $\cdot$ to mean addition also, and since $+$ is taken, they chose $\cdot$. Besides, it even says "Let $(\Bbb Z_6,+)$ and $(\Bbb Z_5,\cdot )$ be two groups". $\endgroup$ – Shaun Nov 21 '19 at 0:29
  • $\begingroup$ Ah, just realized your issue is the operation of $G$ is addition in the first component but multiplication in the second. You're probably fine once you realize this $\endgroup$ – oshill Nov 21 '19 at 0:29
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    $\begingroup$ Yes $(\Bbb Z_5, \cdot)$ is not a group so we cannot define the product ...However $(\Bbb Z_5 \setminus \{0\}, \odot)$ is a group $\endgroup$ – Chinnapparaj R Nov 21 '19 at 0:29
  • $\begingroup$ @Shaun I don't think so; the author defines (\cdot) as a multiplication (saw in other exercises; this is one of them). $\endgroup$ – manooooh Nov 21 '19 at 0:34
  • $\begingroup$ @oshill yeah! The problem is in the second component! Because all the elements of the form $(x,0)$ (where $x\in\Bbb{Z}_6$) does not have inverse! $\endgroup$ – manooooh Nov 21 '19 at 0:35

Assuming $(\Bbb Z_5,\cdot)=U(5)$ is the group of units modulo five, the identity of $G:=(\Bbb Z_6, +)\times U(5)$ is $$([0]_6, [1]_5)$$ and the inverse of $([a]_6, [b]_5)$ in $G$ is $([-a]_6, [c]_5)$, where $c$ is such that $5x+bc=1$ for some $x\in \Bbb Z$ and $$[d]_n:=\{k\in \Bbb Z\mid n \text{ divides } (k-d)\}.$$

Assuming $(\Bbb Z_6,+)=(\Bbb Z_6, +_6)$ and $(\Bbb Z_5,\cdot)=(\Bbb Z_5, +_5)$, then the identity of $H:=(\Bbb Z_6, +_6)\times(\Bbb Z_5, +_5)$ is $$([0]_6, [0]_5)$$ and the inverse of $([a]_6, [b]_5)$ in $H$ is $([-a]_6, [-b]_5)$.

NB: Since five is prime, there is only one group of order five up to isomorphism. The question specifies that $(\Bbb Z_5,\cdot)$ is a group.

  • $\begingroup$ Yes $\Bbb{Z}_{5}=\{0,1,2,3,4\}$. What does $5x+bc=1$ have to do with $[d]_n$? Thank you! $\endgroup$ – manooooh Nov 21 '19 at 0:44
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    $\begingroup$ Well, $5x=1-bc$, so $5\mid 1-bc$. Hence $[1]_5=[bc]_5=[b]_5\cdot [c]_5$, so $[b]_5$ has $[c]_5$ as an inverse. $\endgroup$ – Shaun Nov 21 '19 at 0:48
  • $\begingroup$ But where do we put $[d]_n$? $\endgroup$ – manooooh Nov 21 '19 at 0:49
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    $\begingroup$ No worries, @manooooh. Please don't forget to upvote & accept this answer if it works for you. $\endgroup$ – Shaun Nov 21 '19 at 0:54
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    $\begingroup$ But with your definition of inverse I cannot figure what should be the inverse of $(0,0)$. It should be $(0,c)$, where $c$ is such that $5x+0c=5x=1$ for some $x\in\Bbb{Z}$. But $x=1/5\notin\Bbb{Z}$. So $(0,0)$ has no inverse. What am I doing wrong? $\endgroup$ – manooooh Nov 21 '19 at 0:54

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