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Are the groups $\Bbb Z/8\Bbb Z$, $ (\Bbb Z/2\Bbb Z) \times (\Bbb Z/4\Bbb Z)$ and $(\Bbb Z/2\Bbb Z)^3$ isomorphic?

The solutions says that no because $\Bbb Z/8\Bbb Z$ has an element of order $8$ and not the two others and $(\Bbb Z/2\Bbb Z) \times (\Bbb Z/4\Bbb Z)$ an element of order $4$ and not $(\Bbb Z/2\Bbb Z)^3$.

I want to understand:

  1. Why does $\Bbb Z/n\Bbb Z$ have an element of order $n$?
  2. Why the fact that group $G_1$ has an element of order $k$ and not $G_2$ makes them not isomorphic?

Where I am right now:

  1. If $G$ has an element of order $n$ it means that there is $g \in G$ such that $g^n=e$ ($e$ being the identity element). $\Bbb Z/n\Bbb Z$ is the group in which elements with the same congruence modulo $n$ are equal (eg $2 \equiv 7 \pmod 5$). How does it imply that there is an element of order $n$ in $\Bbb Z/n\Bbb Z$?

  2. $G$ and $G'$ are isomorph if there exists a morphism $f$: $G \rightarrow G'$ and $g$: $G' \rightarrow G$ such that $g \circ f = Id_G$ and $f \circ g = Id_{G'}$. How to use it here?

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  • $\begingroup$ $1$ is an element of order $n$ in $\Bbb Z/n\Bbb Z$ $\endgroup$ Jun 17 '18 at 9:42
  • $\begingroup$ Answer to 1. : 1 $\endgroup$
    – user99914
    Jun 17 '18 at 9:42
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The element $1+n\Bbb Z$ of $\Bbb Z/n\Bbb Z$ has order $n$.

Assume $G_1$ has an element $g$ of order $k$. If $\phi\colon G_1\to G_2$ is a homomorphism, then $\phi(g)^k=\phi(g^k)=\phi(1)=1$, i.e., the order of $\phi(g)$ is at most $k$. If there is an inverse homomoprhism $\psi\colon G_2\to G_1$ then by the same reasoning, the order of $\psi(\phi(g))$ is at most the order of $\phi(g)$. As $\psi(\phi(g)=g$, we conclude that the order of $\phi(g)$ equals $k$.

Loosely speaking, anything that can be "reasonably" stated about a group also holds true for any group isomorphic to it. Where "reasonably" means that the property is expressible solely in terms of group operations ("has an element of order $k$" is such a property, obviously).

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Why does $\mathbb Z / n \mathbb Z$ has an element of order $n$?

The element $\bar{1}$ is an element of order $n$ as $n \bar{1} = \bar{n} = \bar{0}$

Why the fact that group $G_1$ has an element of order $k$ and not $G_2$ makes them not isomorphic?

Suppose that $\phi : G_1 \to G_2$ was an isomorphism between the two groups and $x \in G_1$ is of order $k$. Then $\phi(x)$ is an element of order $k$ in $G_2$. Hence if $G_1$ has an element of order $k$ and not $G_2$, the two groups can't be isomorphic.

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  • $\begingroup$ But then 1 is of order k in Z/nZ for all n and k integers as $1^k \equiv 1 \pmod n$ so 1 is of order 8 in Z/8Z and in Z/2Z x Z/4Z $\endgroup$ Jun 17 '18 at 10:20
  • $\begingroup$ @Arthurim You have to remember the definition of the order of an element. The order of an element $a$ of a group is the smallest positive integer $a$ such that $a^m=e$ $\endgroup$ Jun 17 '18 at 10:23
  • $\begingroup$ True, mea culpa. But then the order of $\bar{1}$ is 0 as $\bar{1}^0≡1^0≡1(\mod n)$. $\endgroup$ Jun 17 '18 at 10:44
  • $\begingroup$ Ok, I just understood my mistake. To be the $Z/nZ$ groups were multiplicative and not additive. Therefore I thought the identity was 1 and not 0. So indeed $\bar{1}$ is of order $n$ in $Z/nZ$ and therefore there is an element of order 8 in Z/8Z and not in Z/2Z x Z/4Z. $\endgroup$ Jun 17 '18 at 13:21
  • $\begingroup$ @mathcounterexamples.net You forgot to prove that $n$ is the smallest positive integer such that $n\cdot\bar 1=\bar 0$. $\endgroup$
    – user26857
    Jun 18 '18 at 9:12
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We say that a group $G$ has exponent $m$ when $g^m=1$ for all $g \in G$. This property is preserved by isomorphisms and it suffices to prove that the groups in question are not isomorphic:

$(\Bbb Z/2\Bbb Z)^3$ has exponent $2$.

$ \Bbb Z/2\Bbb Z \times \Bbb Z/4\Bbb Z$ has exponent $4$ but not $2$.

$\Bbb Z/8\Bbb Z$ has exponent $8$ but not $2$ or $4$.

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