Exercise 67.2

Question

Show that if $G_1$ is a subgroup of $G$, there may be no subgroup $G_2$ of $G$ such that $G = G_1 \oplus G_2$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Let $G = \mathbb{Z}$ and $G_1 = 2\mathbb{Z}$, and suppose that such a $G_2$ exists. Then, we have $G = 2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ by Corollary $67.3$, and so $(0,1) \in G_1 \oplus G_2$. But since $2(0,1) = (0,0)$, $G_1 \oplus G_2$ contains an element of order $2$ while $G$ does not, a contradiction.
\end{proof}

Exercise 67.2

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Exercise 67.2

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