I am asked to express a group G={1,7,17,23,49,55,65,71} under multiplication modulo 96 as an external and internal direct product of cyclic groups. However I also have an example to help with it, but I dont understand how my book is getting their answer.
$\rightarrow$.
The question is
let G={1,8,17,19,26,28,37,44,46,53,62,64,71,73,82,89,91,107,116,127,134} under
multiplication modulo 135. Express G as external and internal direct product of cyclic
group.
Their answer: Since G has order 24, it is isomorphic to one of
$Z_8+Z_3≈Z_{24}$, where + is the direct product
$Z_4+Z_2+Z_3≈Z_{12}+Z_2$
$Z_2+Z_2+Z_2+Z_3≈Z_6+Z_2+Z_2$.
Consider the element 8. Direct calculations show that $8^6=109$ and $8^12=1$. But now we know $G$. Clearly |8|=12 rules out the third group in the list. At the same time, |109|=2=|134|(remember $134=-1\space mod135)$ implies that G is not $Z_{24}$. Thus G must me the middle group and the external product is $<8>$x$<134>$.
I first dont really understand how they came up with all that. Second off, Since my orginal group G has order 8, it cant be re-written as an isomorphism to something because the factors of 8 are not coprime. Where do I go?
Best Answer
Hint. Your group is abelian, and there are three abelian groups of order $8$: $C_2\times C_2\times C_2$, $C_4\times C_2$ and $C_8$, so it must be among them. These are easily distinguished by looking at the orders of elements (as in the example you showed). For instance, if your group has an element of order $8$, then it must be the cyclic group $C_8$. (Further Hint. It's not $C_8$.)