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Suppose $G$ is a group of order $2p$, where $p$ is a prime. Show that there exists an onto homomorphism $G \rightarrow H$, where $H$ is a group of order $2$.

Here is my attempt:

By Cauchy's Theorem, there exists $a \in G$ with order $p$, so $\langle a \rangle \leq G$. Write $H = \langle h \rangle = \{e_H, h\}$ and define $f: G \mapsto H$ by \begin{align*} f(g) = \begin{cases} e_H & \text{if $g \in \langle a \rangle$} \\ h & \text{otherwise} \end{cases} \end{align*}

Is this correct? I am able to show that $f$ behaves like a homomorphism for all cases except when $g_1, g_2 \in G - \langle a \rangle$. That is, I want to show that $g_1g_2 \in \langle a \rangle$ in order for this to be a homomorphism, but I don't know how. Perhaps this is not even the right answer though...

b_pcakes
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1 Answers1

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Hint: By Cauchy's theorem, $G$ has a subgroup of order $p$, which has index $2$ in $G$. Now use the well-known result that a subgroup of index $2$ is normal.

Stahl
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  • So we take $G / \langle a \rangle$ and see that the cosets are $\langle a \rangle$ and $G - \langle a \rangle$, so $G - \langle a \rangle$ is closed under multiplication? – b_pcakes Nov 14 '16 at 07:01
  • The cosets will be $\langle a\rangle$ and $g\langle a\rangle$, where $g\not\in\langle a\rangle$, and you have a natural surjection $G\to G/\langle a\rangle$. – Stahl Nov 14 '16 at 07:03