# Exercise problem in group order & homomorphism

I came across this problem in class note but I am stuck:

Let G be a group of order 21, let G' be a group of order 35, and let φ be a homomorphism from G to G'. Assume that G does not have a normal subgroup of order 3. Show that φ(g) = 1 for each element g in G.

Any help or hints would be very much appreciated. Thanks for your time.

• You mean subgroup of order 7? – Test123 May 14 '14 at 16:06
• @Test123: I will email my professor if there is any typo. Order of 7 does make more sense. Thanks. – Amanda.M May 16 '14 at 14:36
• @A.Mangus It would have been easier but still as you can see below it works fine. – Test123 May 16 '14 at 14:42

The image of $\varphi$ is a subgroup of $G'$ and so its order divides $35$. Moreover, this order is the index of the kernel of $\varphi$ in $G$, and so its order divides $21$. Hence, the order of the image of $\varphi$ is a common divisor of $35$ and $21$ and so divides $7$. But the order of the image cannot be $7$ because otherwise the order of the kernel would be $3$, and $G$ has no normal subgroup of order $3$. Thus, the order of the image of $\varphi$ is $1$.
• @A.Magnus, the image $\varphi(G)$ is isomorphic to the quotient $G/\ker\varphi$. – lhf May 16 '14 at 2:21
If $\varphi\colon G\to G'$ is a homomorphism, then $\ker\varphi$ is a normal subgroup of $G$, so its order is $1$, $3$, $7$ or $21$. It can't be $3$ by assumption. The homomorphism theorem says $\varphi$ induces an injective homomorphism $$\tilde\varphi\colon G/\ker\varphi\to G'$$ such that $\operatorname{im}\tilde\varphi=\operatorname{im}\varphi$.
Call $H$ the image of $\varphi$; what order can $H$ have? Can $\ker\phi$ have order $1$ or $7$?