Prove isomorphism of group

Question

Suppose that $G$ is a finite abelian group and $G$ has no element of order $2$. Is the mapping $f:G \to G$ defined by $f(x) = x^2$ is a group isomorphism?

Answer 1

Yes, it is an automorphism. First note that the map $x\mapsto x^2$ is always an endomorphism if the group is Abelian. Second, the kernel of that map in any Abelian group consists of elements of order $2$ or $1$. Since your group has no elements of order $2$, the kernel is trivial. Since the group is finite, the map is both surjective and injective. So it is an automorphism.

Answer 2

First, it is a homomorphism. That's where the assumption that the group is abelian comes in. (Check this.)

The kernel, under your assumption, is automatically trivial.

Since the group is finite, this means $\varphi$ is surjective.

Thus it's an isomorphism of $G$ with itself, or, an automorphism.

Answer 3

Note that

$f(xy) = (xy)^2 = (xy)(xy) = x(yx)y = x(xy)y = x^2y^2 = f(x)f(y), \tag 1$

that is,

$f:G \to G \tag 2$

is a group homomorphism. We have

$\ker f = \{x \in G \mid x^2 = e \}, \tag 3$

so besides $e$, the identity of $G$, $\ker f$ contains only the elements of order $2$, and since by hypothesis there are no elements of order $2$, it follows that

$\ker f = \{e\}, \tag 4$

and from this follows that $f$ is injective, for if

$x^2 = f(x) = f(y) = y^2, \tag 5$

then

$f(xy^{-1}) = (xy^{-1})^2 = xy^{-1}xy^{-1} = x^2y^{-2} = e, \tag 6$

whence

$xy^{-1} \in \ker f, \tag 7$

so in light of (4),

$xy^{-1} = e, \tag 8$

or

$x = y; \tag 9$

$f$ is thus injective as claimed. Since $G$ is finite, $f$ is also surjective; thus in fact $f$ is an isomorphism from $G$ to itself.

In fact, the preceding demonstration may be extended to cover the case in which $G$ has no elements of order $p$ for some arbitrary prime $p$, and

$f(x) = x^p; \tag{10}$

such an $f$ is clearly a homomorphism:

$f(xy) = (xy)^p = x^py^p = f(x)f(y), \tag{11}$

and in keeping with the above,

$\ker f = \{x \in G \mid x^p = e \}, \tag{12}$

which via our hypothesis implies

$\ker f = \{e\}; \tag{13}$

then as in (5),

$x^p = f(x) = f(y) = y^p, \tag{14}$

and as in (6),

$f(xy^{-1}) = (xy^{-1})^p = x^py^{-p} = e, \tag{15}$

so once again (13) allows us to infer

$xy^{-1} = e, \tag{16}$

or

$x = y, \tag{17}$

and again $f$ is injective, hence (from the finiteness of $G$), an isomorphism.

Note Added in Edit, Thursday 8 October 2020, 10:40 AM PST: We expand upon the inference, made above ca. (12)-(13), that $\ker f$ consists solely of the identity element $G$; for if not, we may find some $y \in G$,

$e \ne y \in \ker f, \tag{18}$

with

$1 < q = o(y) = \text{order}(y) < p \subset G, \tag{19}$

since every element of $\ker f$ has order at most $p$, by definition. (It may help the reader to recall at this point that the order of a group element $y$ is the least positive integer $q$ such that $y^q = e$.) We show that

$q \mid p; \tag{20}$

now by Euclidean division we may uniquely write

$p = aq + r, \; a, r \in \Bbb Z, \; 0 \le r < q, \tag{21}$

from which

$y^r = ey^r = e^ay^r = (y^q)^ay^r = y^{aq + r} = y^p = e, \tag{22}$

which contradicts (18)-(19) unless $r = 0$, in which case (22) reduces to the triviality

$y^0 = e. \tag{23}$

Thus

$r = 0 \tag{24}$

and

$p = aq, \tag{25}$

whence

$q \mid p \tag{26}$

as claimed; now since $p$ is prime we have

$q = 1 \; \text{or} \; q = p; \tag{27}$

if $q = 1$ then $y = y^q = e$, whilst $q = p$ is ruled out by hypothesis (19) that $q < p$. Thus there is no $y$, $q$ such that (18)-(19) bind, and hence we have (13).

Finally, we observe that the hypothesis $p$ is prime cannot be omitted, for if $p$ is composite, say

$p = uv, \; 1 < u, v < p, \tag{28}$

then it is possible that

$x^u = e, \tag{29}$

whence

$x^p = x^{uv} = (x^u)^v = e^v = e, \tag{30}$

but the order of $x$ is at most $u$; in a case like this, we have

$x \in \ker f, \tag{31}$

but

$o(x) \ne p. \tag{32}$

It may be seen that the case $p = 2$ circumvents the need for these detailed arguments, since the factorization (28) is impossible in this case.

It seems to me to be quite likely that examples illustrating these points may easily be constructed; I leave the details to my extensive readership.

End of Note.