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First of all, I'm new in Group Theory. I'm trying to understand how to determinate certain orders of elements in a group. For example:

Indicate the order of the following elements: $a=35_{42} \in \mathbb{Z}_{42} , \ b=(3_{27},(123)) \in \mathbb{Z}_{27}\times S_5$

I know that the order of an element $x$ of a group G is the lower positive value $k$ such that $x^k=e_G$. Given this, I can compute for the element $a$:

$35^1=35$,

$35^2=28$,

$35^3=21$,

$35^4=14$,

$35^5=7$,

$35^6=42=0=e_G$

So the order of $a$ is $6$. However I don't know how to use this argument to determinate the order of $b=(3_{27},(123)) \in \mathbb{Z}_{27}\times S_5$

2 Answers 2

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You know that $9 \cdot 3\mod 27=0$ and $(123)^3=Id$. Hence $(3,(123))^9=(9\cdot 3,(123)^9)=(0,Id)$ which is the neutral element in $\mathbb{Z}_{27}\times S_5$. So clearly the order of $(3,(123))$ is bounded by $9$. Since $8\cdot 3\mod 27\neq 0$, the order is $9$.

Is there any relation between $9$ and $3$ that springs to mind? There is a general rule for determining the order of elements in direct sums of groups. Try playing around with different orders to find the answer.

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    Is this true : ord(a,b)=LCM(ord(a),ord(b))??? – 2017-01-20
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    Yes, now prove it using the answer by RSerrao – 2017-01-20
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HINT

If an element $b$ is of the form $(b_1, b_2)$ where $b_1, b_2$ are from some groups $G_1, G_2$, then first find the order of $b_1$ in $G_1$ and the order of $b_2$ in $G_2$. Then note that $b^k$ will only be the identity of $G_1\times G_2$ if $b_1^k = e_{G_1}$ and $b_2^k = e_{G_2}$.

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    I didn't quite understand what you said. First I find the order of $b_1$ and $b_2$ and what do I do after that?? – 2017-01-20
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    @AndreMath yes; after you find the orders of $b_1$ and $b_2$ separately, call them $k_1, k_2$, you know that: $(b_1, b_2)^{k_1} = (e_{G_1}, b_2^{k_1})$ and $(b_1, b_2)^{k_2} = (b_1^{k_2}, e_{G_2})$ right? But you need the identity in the two coordinates at the same time. Thus you are looking for a multiple of $k_1$ that is also a multiple of $k_2$. Indeed, you want the smaller number that is multiple of $k_1$ and $k_2$ at the same time. – 2017-01-20