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Sizes of kernels of homomorphisms


Oct 7, 2013
I have a problem that I have been stuck on for two hours. I would like to check if I have made any progress or I am just going in circles.

**Problem: Let $\alpha:G \rightarrow H, \beta:H \rightarrow K$ be group homomorphisms. Which is larger, $\ker(\beta\alpha)$ or $\ker(\alpha)$?**

**My work:** $\ker(\alpha)<G, \ker(\beta\alpha)<G, |G|=|\ker\alpha|*|im\alpha|=|\ker\beta|*|im\beta|$


$|im\alpha|$ divides $|G|$ and $|H|$

$|im\beta\alpha|$ divides $|G|$ and $|K|$

$|ker(\beta\alpha)|=\frac{|G|}{|im(\beta\alpha)|}, |ker(\alpha)|=\frac{|G|}{|im(\alpha)|}$

$\frac{|\ker(\beta\alpha)|}{|\ker\alpha|} \leq \frac{|H|}{|im(\beta\alpha)|}$

With similar analysis, I get $|\ker(\beta\alpha)| \geq \frac{|G|}{|K|}$ and $|\ker(\alpha)| \geq \frac{|G|}{|H|}$.

This seems like too much work with zero output.


Well-known member
MHB Math Scholar
Feb 15, 2012
If $g \in \text{ker}(\alpha)$ then it is immediate that:

$\beta\alpha(g) = \beta(e_H) = e_K$, so that:

$g \in \text{ker}(\beta\alpha)$.

Hence it is obvious that all of $\text{ker}(\alpha)$ is contained within $\text{ker}(\beta\alpha)$, so the latter must be the larger set (it may the same set if $\beta$ is an isomorphism).