Let $p$ be an odd prime. Suppose that $a$ is an odd integer and also $a$ is a primitive root mod $p$. Show that $a$ is also a primitive root mod $2p$. Let $p$ be an odd prime. Suppose that $a$ is an odd integer and also
$a$ is a primitive root modulo $p$. Show that a is also a primitive root modulo
$2p$.
Any hints will be appreciated. Thanks very much.

More generally, if ord$_{m_1}a=d_1$ and ord$_{m_2}a=d_2$  with $(m_1,m_2)=1$
we can prove ord$_{(m_1\cdot m_2)}a=$lcm$(d_1,d_2)$ (Proof below)
Put $m_1=p^k,m_2=2^n$ with prime $p>2$
Then choose $n=1$
Proof: 
ord$_{m_1}a=d_1\implies a^{d_1}\equiv1\pmod{m_1}$
and ord$_{m_2}a=d_2\implies a^{d_2}\equiv1\pmod{m_2}$
If $(d_1,d_2)=g$ and $\dfrac{d_1}{D_1}=\dfrac{d_2}{D_2}=g$ so that $(D_1,D_2)=1$
lcm$(d_1,d_2)=gD_1D_2$ 
Clearly, $a^{gD_1D_2}\equiv1\pmod{m_1}$ and $a^{gD_1D_2}\equiv1\pmod{m_2}$
As $(m_1,m_2)=1,a^{gD_1D_2}\equiv1\pmod{m_1m_2}$
If ord$_{m_1\cdot m_2}a=D,D$ must divide $gD_1D_2\ \ \ \ (1)$
Again, $a^D\equiv1\pmod{m_1m_2}\implies a^D\equiv1\pmod{m_1}\implies d_1(=gD_1)$ must divide $D$
Similarly, $d_2(=gD_2)$ must divide $D$
$\implies$lcm$(gD_1,gD_2)=gD_1D_2$ must divide $D\ \ \ \ (2)$
$(1),(2)\implies D=gD_1D_2$
