How to show the finite normal subgroup of $\Bbb Z_p*\Bbb Z_p*\Bbb Z_p$ is trivial?

How to show the finite normal subgroup of $\Bbb Z_p*\Bbb Z_p*\Bbb Z_p$ is trivial? (Here $p$ is prime.)

Here is my tentative proof:

Represent $\Bbb Z_p*\Bbb Z_p*\Bbb Z_p$ as $\langle a,b,c\mid a^p=b^p=c^p=1\rangle$. If $N$ is a finite subgroup of this group, then to $n\in N$ we may find an element $g_1$ in $\{a,b,c\}$ such that $g_1cg_1^{-1}\in N$ is reduced, and similarly $g_2$ in $\{a,b,c\}$ such that $g_2g_1cg_1^{-1}g_2^{-1}\in N$. We can repeat the procedure to get a sequence of conjugations in $N$.

The thing left is to show they are all different, then $N$ is infinite, thus we get a contradiction. But it's obvious the conjugations in sequence are different to each other, isn't it?

An element $1 \ne g$ can be written in normal form as a product $g_1g_2 \cdots g_k$ for some $k \ge 1$, where each $g_i$ has the form $x^i$ with $x \in \{a,b,c\}$, $1 \le i < p$, and adjacent $g_i$ are powers of different generators.
Suppose WLOG that $g_1$ is a power of $a$. Then for any normal form word $w$ in the infinite group $\langle b,c \rangle$, the term $g_1$ cannot cancel when reducing the word $wgw^{-1}$ to normal form, so its normal form has the prefix $wg$. So they are all distinct and hence $g$ has infinitely many distinct conjugates and cannot lie in a finite normal subgroup.
• Thanks and if I change $w$ to be the normal form word with its last term is in $\langle b,c \rangle$, then it will still be ok right? – Katherine Oct 1 '16 at 22:27
• Not necessarily. You could have $g=ab$ and $w$ any power of $ab$. They all give $wgw^{-1}=ab$. – Derek Holt Oct 2 '16 at 10:47