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I am currently looking back at some exercises of my first classes of algebra. I had this question: Given that $A, \cdot$ is a group of integers modulo 56 and $5, 15 \in A$. What are the other elements?

So i have tried to solve this using what we saw up to that point of the lectures: the definition of groups and the definition of subgroups (+ subgroup criterion).

I'm kind of stuck on this question: I started multiplying 5 with itself, so I have found that $5^2 = 25 \in A, 5^3 = 125 = 13 \in A$ and $5^4 = 13 \cdot 5 = 9 \in A$... Then I looked at $5 \cdot 15 = 75 = 19$. After this, I made some more computations to discover new elements at each try.

However, there must be a quicker way, right?

Based on the fact that $\text{gcd}(5,56) = 1$, I know that (using more information than I knew at that time) that $5$ is a unit (same thing for 15).

Any hints on how to solve this (using only definition of groups + definition of subgroups + groupcriterion)?

Thank you in advance.

EDIT: I have found (by just computing products, that $5^6 = 1 (\mod 56)$.

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    There are structural results for groups of this form, but they get more complicated because $56$ is not prime. It might be easier to just write out elements. – 2017-02-07
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    so I just have to compute all possible products of elements I find? Since at this point, I already have 6 different elements and because of the $\text{gcd}$ I think I might find some more, which leads to a lot of products to compute. This made me think that there must be some shorter way... – 2017-02-07
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    A few quick thoughts: there can be no even integers in the subgroup because $5$ and $15$ are both odd, and $56$ is even. That restricts the size of the group. Also, $15 = 3\cdot5$ so any multiple of $15$ is automatically a multiple of $5$ so I think you only need to look at powers of $5$. And I think it's Femat's little theorem that will tell you which $n$ is such that $5^n= 1 \mod 56$ which is the last power you need to calculate – 2017-02-07
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    just compute $15 \cdot 15$ which appears to be $1$ in this group. So I guess you are right :) thank you! – 2017-02-07

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Note that $15^2 \equiv 1$ (all congruences modulo $56$).

Note that the powers of $5$ are $$ \begin{cases} 5^0 \equiv 1,\\ 5^1 \equiv 5,\\ 5^2 \equiv 25,\\ 5^3 \equiv 13,\\ 5^4 \equiv 9,\\ 5^5 \equiv 45,\\ 5^6 \equiv 1. \end{cases}$$ So the group generated by $5$ and $15$ consists of the $2 \cdot 6 = 12$ elements $$ 5^x \cdot 15^y $$ where $0 \le x < 6$ and $0 \le y < 2$. Try and show these elements are indeed distinct.

Note that the formulation of the problem is somewhat ambiguous, because the group of invertible elements modulo $56$, of order $24$ also contains the two given elements.

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    I suppose that I need to find the smallest subgroup, but this could have been written in the question I guess. Thank you for your quick answer! – 2017-02-07
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    @Student, thanks! – 2017-02-07
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    @Student: The group $A \cong C_2 \times C_2 \times C_6$ so you need at least three generators. – 2017-02-07
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    @marcbogaerts: Why two times $C_2$? I have only got two elements and was asked to give al elements of the group $A$... – 2017-02-07
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    @Student: Good question. Normally the units of $\Bbb Z_{p^k}$ form a cyclic group of order $p^{k-1}(p-1)$. An exception to this arises for $p = 2$ and $k \geq 3$. It is not hard to verify by hand that the units of $\Bbb Z_8$ is the group $C_2 \times C_2$ instead of $C_4$ as it should be expected to be. You can find a reference to an online article with proof in my answer to [this question](http://math.stackexchange.com/questions/2130763/structure-of-z-nz/2131778#2131778). Maybe a good question for this site? – 2017-02-07
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    But i just don't see why this comment is relevant to this answer to my question... I found that A has 12 elements using this answer... This question was An exercise in my algebra course, where i only saw groups and subgroups... How do you see that, given $2$ elements in $A$, it should have three generators? Of is this described in your answer? – 2017-02-07
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    @Student: You only have to look at the order of the units of $\Bbb Z_8$,, none of them have order $4$ . – 2017-02-07
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    How does $\mathbb{Z}_8$ enters this discussion? Sorry, i am really confused. – 2017-02-07
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    @marc Bogaerts: are you suggesting that the answer given by Andreas caranti is wrong? Or is this related to his comment about the group of order 24 also being a possibility? – 2017-02-07
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    $\Bbb Z_8$ because $56 = 8*7$ and $U(mn) = U(m)U(n)$ if $\gcd(m,n) = 1$. The answer of Andreas is completely correct and he correctly remarks that the group $A$ has only $12$ elements while the group of all units of $\Bbb Z_{56}$ has $24$ elements.If you don't understand why $U(56) = U(8)\times U(7)$ you should put it as a separate question because an answer in the comments is too long. – 2017-02-07
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    oh, but this I understand! I was really bothered that I did not understand what could be possibly wrong with the answer of Andreas. However, I see your remark about the units, and I think I understand this. Thank you! – 2017-02-07