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I have two questions, the type of which I always struggle to answer. These are questions from some old papers. I've tried to find out from my textbook how to handle these $x^a = e$ questions, but I guess I'm misreading it, because I just can't seem to find any info on it.

How many abelian groups $G$ of order 256 are there (up to isomorphism) with the property that $x^4 = e$ for all $x \in G$?

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True or False, and explain: If $G$ is a group of order 8, and not cyclic, then $x^4 = 1$ for all $x \in G$.

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    If it existes $x \in G$ such that $x^4 \neq 1$, you should consider the group generated by $x$. For your first question, you can use the classification theorem about abelian groups.2012-05-11
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    Hi, could you please elaborate on that?2012-05-11
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    Just to be precise I would add to the comment of Monoide that your group is finite *and* abelian...2012-05-11
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    The classification theorem for abelian and finite groups states that every finite abelian group is a direct product of cyclics groups. For exemple in your case, $(\mathbb{Z}/2\mathbb{Z})^8$ and $(\mathbb{Z} / 4\mathbb{Z}) \times (\mathbb{Z} / 2 \mathbb{Z})^6$ are two non isomorphic abelians groups of order 256.2012-05-11
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    Do you mean I should factor 256 into primes? $2^8 = 256$ ?2012-05-11
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    Exactly, in order to apply the classification theorem.2012-05-11
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    So should I find all the different combinations of factoring 256 into its prime components? Is that the number of abelian groups? I still don't understand how $x^4 = e$ fits into this.2012-05-11
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    You have the one prime 2, and you can split your group into factors which are powers of 2 - cyclic groups of order 2,4,8,16,32,64,128,256. If one of your factors has order 8 or above, it will have an element of order 8. So you need to avoid those groups and find the decompositions which do not include them.2012-05-11
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    There's only one way to factoring an integer into prime components... What you are looking for is the number of way to factor 256. See the theorm on Wikipedia for a precise statement. When you will have all abelian groups of order 256, the (easy) point is to see whose of them satisfy your condition, or not.2012-05-11
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    Okay, I think I'm starting to understand better. So, Mark, what you're saying is in a group of order 8 or greater, an element $x$ of order $4$ would not complete its cycle back to $e$. Is that it?2012-05-11

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I will start from the second part, as it is easier. You use Lagrange's theorem that the order of an element $x$ divides the order of the group. In a group of order 8, therefore, either $\operatorname{ord} x=8$ or it divides $4$. In the former case, as @Monoide writes, the group coincides with the cyclic group generated by $x$. In the latter, since $4$ is a multiple of $\operatorname{ord} x$ we have $x^4=1$.

Now the same reasoning can be applied in the first case. Note that $256=2^8$. By the classification theorem for abelian groups, $G$ is isomorphic to a direct sum of cyclic groups of order dividing $2^8$. If we had a cyclic group of order $2^n$ with $n\geq 3$ in that direct sum, then we would have an element of order $2^n>4$ in $G$, against your assumption. So in the end, we are left to count the number of ways to write $256$ as an unordered product of $2$ and $4$'s. Taking logarithms to the base 2, this is the same as writing $8$ as an unordered sum of $1$ and $2$'s. You can do this by counting the number of $2$'s in this sum (the other terms then being $1$). Since there are anywhere less than or equal to four $2$'s, the answer is five ($0,1,2,3,4$).

In general, the answer to the same question for any abelian group $G$ of order $2^n$ is $1+\lfloor n/2 \rfloor$.

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    You forgot $(\mathbb{Z}/2\mathbb{Z})^n$, so the general answer is $1 + \lfloor n/2 \rfloor$. In this particular exemple, it is $5$.2012-05-11
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    You're right, I forgot when there are no $2$'s at all. I will edit the answer. Thanks!2012-05-11
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    Clear and well-done.2012-05-11