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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 $x \in G$?


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. – Monoide May 11 '12 at 12:13
Hi, could you please elaborate on that? – janvdl May 11 '12 at 12:28
Just to be precise I would add to the comment of Monoide that your group is finite and abelian... – Giovanni De Gaetano May 11 '12 at 12:29
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. – Monoide May 11 '12 at 12:38
Do you mean I should factor 256 into primes? $2^8 = 256$ ? – janvdl May 11 '12 at 12:40
up vote 3 down vote accepted

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 $5$. – Monoide May 11 '12 at 13:09
You're right, I forgot when there are no $2$'s at all. I will edit the answer. Thanks! – Francesco Sica May 11 '12 at 13:11
Clear and well-done. – awllower May 11 '12 at 15:46

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