# group $G$ such that $|G| = p^k$ for some prime $p$ and some integer $k > 1$, where $|g| = p$ for every $g \in G \setminus {\epsilon}$

Question:
Give an example of a group $$G$$ such that $$|G| = p^k$$ for some prime $$p$$ and some integer $$k > 1$$, where $$|g| = p$$ for every $$g \in G \setminus {\epsilon}$$.
My Reasoning :
I wasn't too sure how to approach this question. Would the group $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ be such a group? I wasn't too sure because the operation of this group is addition and not multiplication so I can't take powers of elements, which is what this question asks. for example : $$(1,0)^2 = (1,0)$$ $$\neq$$ $$(0,0)$$. I also thought that I can just take the general Klein group where: $$V = \{1, a,b,c\mid a^2=b^2=c^2\}$$. Are these two groups good examples? if not, what other group can I pick?

• If you use multiplicative notation, $a^n$ is a shorthand for an n-fold product $a \cdots a$. If you use additive notation, you write $n\cdot a$, which then means $a + \cdots + a$ ($n$ times). If the problem statement uses multiplicative notation, it does not restrict you to it - use can use additive notation too, just translate carefully products to sums and powers to products. Commented Mar 25, 2023 at 19:00
• If you follow ALG's remark, both your examples are valid for your problem. [Note also that the two examples you give are isomorphic groups, just keep in mind the operations are multiplication in one of them and addition in the other. Commented Mar 25, 2023 at 19:16
• $g^n$ just means "the group operation being applied to $g$ $n$ times"; or, more visually, $\underbrace{g\cdot g\cdot ...\cdot g}_{n\text{ times}}$. It doesn't matter if the group operation is addition or multiplication or something else entirely.
– IAAW
Commented Mar 25, 2023 at 20:31
• Those two groups are actually... one, i.e. they are isomorphic. More generally, $C_p\times C_p$ has that property, for every prime $p$. Commented Mar 25, 2023 at 21:50
• To make this more interesting, try to find a non-abelian group with this property. Commented Mar 26, 2023 at 7:53

More generally: $$G=C_p\times C_p$$ where $$C_p$$ is the cyclic group of order $$p$$ (in multiplicative notation), has those properties: it has order $$p^2$$ and every nontrivial element has order $$p$$. Your example (one, actually, as $$\Bbb Z_2\times \Bbb Z_2\cong V$$) is just the special case for $$p=2$$. So, yes, it is a good example.