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The only such possible group is $V$ (up to isomorphism). If $\phi$ be such an into homomorphism, then $o(\phi(V))=4$ and $\phi(V)$ being a subgroup of $\mathbb{Z}_8$, it must be cyclic with a generator of order $4$. But, $V$ has not element of order $4$. A contradiction.

Does this work?

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  • 1
    $\begingroup$ What is an into homomorphism? $\endgroup$ – José Carlos Santos Apr 17 at 10:36
  • $\begingroup$ A monomorphism. $\endgroup$ – Subhasis Biswas Apr 17 at 10:37
  • $\begingroup$ i.e. $\forall y \in \phi(G)$, $\{x \in G: \phi(x)=y\}$ has only one element. $\endgroup$ – Subhasis Biswas Apr 17 at 10:40
  • $\begingroup$ It is a nice exercise to prove that every subgroup of a cyclic group is cyclic. Your result follows immediately. $\endgroup$ – user1729 Apr 17 at 11:16
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That argument works. Given a monomorphism $f:G\to H$ between groups $G,H$, then $G$ is isomorphic to $f(G)$, and the two must therefore have the same number of elements of any given order.

Alternatively, there is only one element of order $2$ in $\Bbb Z_8$, while there are three in $V$.

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  • $\begingroup$ Nice argument! Can you please post some links here to very similar group theory questions? (Contest math type). Thanks! $\endgroup$ – Subhasis Biswas Apr 17 at 11:01
  • $\begingroup$ @SubhasisBiswas Any decent introductory group theory textbook should have lots of problems just like this one. $\endgroup$ – Arthur Apr 17 at 11:02
  • $\begingroup$ I am going through. $\endgroup$ – Subhasis Biswas Apr 17 at 11:05

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